[PHP-DB] please help..new problems php. 5.5.7.tar.bz2 and iis 5.1

[Fred Silsbee]

downloaded php 5.5.7.tar.bz2

see attached screenshot

I did an "Add" and the dialog requested a suffix (php) and a filename

where is the file name and where might it be found

I did this same task about 3 years ago and it worked 
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Re: [PHP-DB] Please help a newbie

[mrfroasty]

Thanks
Great tip there...

Gr
mrfroasty


Chris wrote:

mrfroasty wrote:

Hello,

May be try something like this:

$query1=
CREATE TABLE contacts(
id int(16) NOT NULL auto_increment,
phone varchar(15)  NOT NULL,
name varchar(15) NOT NULL,
address varchar(15) NOT NULL,
PRIMARY KEY (id)
);

$query2 = "INSERT INTO contacts VALUES ('NULL','$phone', '$name', 
'$address')";


P:S
id incremented automatically by MYSQL now


Maybe - but it's by accident. You're trying to insert the word NULL 
into an int field (it's being treated as a word because of the single 
quotes around it).


Don't specify the id field at all:

$query2 = "insert into contacts(phone, name, address) values ('" . 
mysql_real_escape_string($_POST['phone']) . "', '" . 
mysql_real_escape_string($_POST['name']) . "', '" . 
mysql_real_escape_string($_POST['address']) . "')";


You should always use the field names (as above) because if your table 
gets reordered, your inserts will now break - if you put "name" before 
phone, the data is now going into the wrong fields.





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Re: [PHP-DB] Please help a newbie

[Chris]

mrfroasty wrote:

Hello,

May be try something like this:

$query1=
CREATE TABLE contacts(
id int(16) NOT NULL auto_increment,
phone varchar(15)  NOT NULL,
name varchar(15) NOT NULL,
address varchar(15) NOT NULL,
PRIMARY KEY (id)
);

$query2 = "INSERT INTO contacts VALUES ('NULL','$phone', '$name', 
'$address')";


P:S
id incremented automatically by MYSQL now


Maybe - but it's by accident. You're trying to insert the word NULL into 
an int field (it's being treated as a word because of the single quotes 
around it).


Don't specify the id field at all:

$query2 = "insert into contacts(phone, name, address) values ('" . 
mysql_real_escape_string($_POST['phone']) . "', '" . 
mysql_real_escape_string($_POST['name']) . "', '" . 
mysql_real_escape_string($_POST['address']) . "')";


You should always use the field names (as above) because if your table 
gets reordered, your inserts will now break - if you put "name" before 
phone, the data is now going into the wrong fields.


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Re: [PHP-DB] Please help a newbie: Fixed

[Rij]

Ok, figured it out.
I was using an int for the field but the value was too large for it. I
used bigint instead and life is suddenly so much better. Well it helps
that its bright and sunny here too :)

On Sun, Apr 19, 2009 at 10:37 AM, mrfroasty  wrote:
> Try using var_export() or var_dump() to debug to see why that 1st time is
> having different values than what you want :-)
> The idea from the sample code I have provided is not to use phone as id...
>
> P:S
> A wild guess would phone is some sort of string.
>
> GR
> mrfroasty
>
>
>
> Rij wrote:
>>
>> Ok. But that still doesn't tell me why I am getting the behavior from
>> my code. Where is the garbage value coming from? And why only the
>> first time I do an INSERT?
>>
>> To Daniel Carrera, thanks for your tip. I sure will look up your
>> suggestion.
>>
>> On Sun, Apr 19, 2009 at 2:32 AM, mrfroasty  wrote:
>>
>>>
>>> Hello,
>>>
>>> May be try something like this:
>>>
>>> $query1=
>>> CREATE TABLE contacts(
>>> id int(16) NOT NULL auto_increment,
>>> phone varchar(15)  NOT NULL,
>>> name varchar(15) NOT NULL,
>>> address varchar(15) NOT NULL,
>>> PRIMARY KEY (id)
>>> );
>>>
>>> $query2 = "INSERT INTO contacts VALUES ('NULL','$phone', '$name',
>>> '$address')";
>>>
>>> P:S
>>> id incremented automatically by MYSQL now
>>>
>>> GR
>>> mrfroasty
>>>
>>>
>>>
>>>
>>
>>
>
>
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>

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Re: [PHP-DB] Please help a newbie

[Daniel Carrera]

Rij wrote:

I input the values from a HTML form. Here is the partial code.
$phone = $_POST['phone'];
$name  = $_POST['name'];
$address = $_POST['address'];
$query = "INSERT INTO contacts VALUES ('$phone', '$name', '$address')";
if (mysql_query($query, $con)) echo "Values inserted";
else die('Unable to create table : '.mysql_error());


This is unsafe code. I suggest you lookup "prepared statements" and the 
PDO library (which is part of PHP).


Daniel.

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Re: [PHP-DB] Please help a newbie

[mrfroasty]

Hello,

May be try something like this:

$query1=
CREATE TABLE contacts(
id int(16) NOT NULL auto_increment,
phone varchar(15)  NOT NULL,
name varchar(15) NOT NULL,
address varchar(15) NOT NULL,
PRIMARY KEY (id)
);

$query2 = "INSERT INTO contacts VALUES ('NULL','$phone', '$name', 
'$address')";


P:S
id incremented automatically by MYSQL now

GR
mrfroasty



Rij wrote:

Hello,

I am new to the world of PHP and MySQL. My objective is to create a
table, insert values in it and read it back.

Here's the partial code to create a table from a PHP file:

if (!$table_exists) {
$query="CREATE TABLE contacts (id int(20) NOT NULL, name
varchar(15) NOT NULL, address varchar(15),PRIMARY KEY(id)
if (mysql_query($query, $con)) echo "Table contacts created";
else die('Unable to create table : '.mysql_error());
}


I input the values from a HTML form. Here is the partial code.
$phone = $_POST['phone'];
$name  = $_POST['name'];
$address = $_POST['address'];
$query = "INSERT INTO contacts VALUES ('$phone', '$name', '$address')";
if (mysql_query($query, $con)) echo "Values inserted";
else die('Unable to create table : '.mysql_error());


Now the problem that I am facing is that when I make my first insert,
the id field shows a garbage value and not the number that I entered.
Subsequent entries into the table show up just fine. It's only the
first one.

What am I doing wrong?

Thanks, Rij

  



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[PHP-DB] Please help a newbie

[Rij]

Hello,

I am new to the world of PHP and MySQL. My objective is to create a
table, insert values in it and read it back.

Here's the partial code to create a table from a PHP file:

if (!$table_exists) {
$query="CREATE TABLE contacts (id int(20) NOT NULL, name
varchar(15) NOT NULL, address varchar(15),PRIMARY KEY(id)
if (mysql_query($query, $con)) echo "Table contacts created";
else die('Unable to create table : '.mysql_error());
}


I input the values from a HTML form. Here is the partial code.
$phone = $_POST['phone'];
$name  = $_POST['name'];
$address = $_POST['address'];
$query = "INSERT INTO contacts VALUES ('$phone', '$name', '$address')";
if (mysql_query($query, $con)) echo "Values inserted";
else die('Unable to create table : '.mysql_error());


Now the problem that I am facing is that when I make my first insert,
the id field shows a garbage value and not the number that I entered.
Subsequent entries into the table show up just fine. It's only the
first one.

What am I doing wrong?

Thanks, Rij

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Re: [PHP-DB] Please help

[Micah Stevens]

WHERE columname LIKE condition

^^^ example.





On Friday 10 September 2004 10:22 am, Stuart Felenstein wrote:
> Okay, I was under the impression that "where" is
> implied inthe joins xx.xx = xx.xx .  Is that not the
> case ?
> As a matter of fact, there isn't even a where or like
> in my query.  Then again myabe that's why I can see
> *all* records but not run a search.
>
> Can you please provide an exmample of the column name
> bteween where and like ?
>
> Thank you.
> Stuart
>
> --- Micah Stevens <[EMAIL PROTECTED]> wrote:
> > If you look at the query, there's no column name
> > between 'WHERE' and 'LIKE'
> > which is a syntax error. That's the problem.
> >
> >
> > Looks like your sql generator has some issues, or
> > you didn't specify the WHERE
> > column properly. I'm not familiar with the system
> > you're using, but keep in
> > mind, that mysql_error will usually not steer you
> > wrong.
> >
> > -Micah
> >
> > On Friday 10 September 2004 10:03 am, Stuart
> >
> > Felenstein wrote:
> > > I had not, my apologies.  I think your post
> >
> > slipped
> >
> > > by.  Anyway, yes I have now inserted mysql_error()
> >
> > and
> >
> > > got my return.  Though I'm not entirely sure how
> >
> > to
> >
> > > fix it.
> > > Here is the error.
> > >
> > > You have an error in your SQL syntax. Check the
> >
> > manual
> >
> > > that corresponds to your MySQL server version for
> >
> > the
> >
> > > right syntax to use near 'Like 'ACCFIN'' at line 1
> > > SELECT `VendorJobs`.`JobID`,
> > > `VendorSignUp`.`CompanyName`,
> > > `StaIndTypes`.`CareerCategories`,
> > > `StaUSCities`.`City`, `USStates`.`States`,
> > > `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`,
> > > `VendorJobs`.`PayRate`,
> >
> > `staTravelReq`.`TravelReq`,
> >
> > > `VendorJobDetails`.`Details`,
> > > `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle`
> >
> > FROM
> >
> > > `VendorJobs` INNER JOIN `VendorSignUp` ON
> > > (`VendorJobs`.`VendorID` =
> >
> > `VendorSignUp`.`VendorID`)
> >
> > > INNER JOIN `StaIndTypes` ON
> >
> > (`VendorJobs`.`Industry` =
> >
> > > `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN
> > > `StaUSCities` ON (`VendorJobs`.`LocationCity` =
> > > `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates`
> >
> > ON
> >
> > > (`VendorJobs`.`LocationState` =
> >
> > `USStates`.`StateID`)
> >
> > > LEFT OUTER JOIN `staTaxTerm` ON
> > > (`VendorJobs`.`TaxTerm` =
> >
> > `staTaxTerm`.`TaxTermID`)
> >
> > > INNER JOIN `staTravelReq` ON
> >
> > (`VendorJobs`.`TravelReq`
> >
> > > = `staTravelReq`.`TravelReqID`) INNER JOIN
> > > `VendorJobDetails` ON (`VendorJobs`.`JobID` =
> > > `VendorJobDetails`.`JobID`) where Like 'ACCFIN'
> >
> > limit
> >
> > > 0,1
> > >
> > > --
> > >
> > > --- Micah Stevens <[EMAIL PROTECTED]>
> >
> > wrote:
> > > > did you make the change to the code I suggested?
> > > > What does MySQL say the error
> > > > is?
> > > >
> > > > -Micah
> > > >
> > > >
> > > > On Friday 10 September 2004 07:49 am, Stuart
> > > >
> > > > Felenstein wrote:
> > > > > As I said this is a code generator
> >
> > (dbqwiksite).
> >
> > > > So,
> > > >
> > > > > describing the process for creating the code
> >
> > is
> >
> > > > > different.  The $sql is fine, as far as typos
> >
> > or
> >
> > > > > incorrect characterrs.  I've gone through
> >
> > those
> >
> > > > > statement very carefully.
> > > > > I've also tried to run a debug with no luck.
> > > > > But I do know that there is something wrong
> >
> > with
> >
> > > > the
> > > >
> > > > > statement, since I can get a complete display
> >
> > of
> >
> > > > all
> > > >
> > > > > records, but can't search using criteria.
> > > > >
> > > > > Below is the $sql, one I know works , though
> >
> > I've
> >
> > > > > tried others.  VendorJobs is the table I'm
> > > >
> > > > querying ,
> > > >
> > > > > and since it's made up of values from other
> >
> > static
> >
> > > > > type tables I have all the joins in place.
> > > > >
> > > > > I've also taken the tick marks out, thinking
> >
> > maybe
> >
> > > > it
> > > >
> > > > > was a style issue.  I'm not sure where to go
> >
> > with
> >
> > > > it.
> > > >
> > > > > Getting a response from the company is like
> >
> > pretty
> >
> > > > > difficult.
> > > > >
> > > > >   `VendorJobs`.`JobID`,
> > > > >   `VendorSignUp`.`CompanyName`,
> > > > >   `StaIndTypes`.`CareerCategories`,
> > > > >   `StaUSCities`.`City`,
> > > > >   `USStates`.`States`,
> > > > >   `VendorJobs`.`AreaCode`,
> > > > >   `staTaxTerm`.`TaxTerm`,
> > > > >   `VendorJobs`.`PayRate`,
> > > > >   `staTravelReq`.`TravelReq`,
> > > > >   `VendorJobDetails`.`Details`,
> > > > >   `VendorJobs`.`PostStart`,
> > > > >   `VendorJobs`.`JobTitle`
> > > > > FROM
> > > > >   `VendorJobs`
> > > > >   INNER JOIN `VendorSignUp` ON
> > > > > (`VendorJobs`.`VendorID` =
> > > >
> > > > `VendorSignUp`.`VendorID`)
> > > >
> > > > >   INNER JOIN `StaIndTypes` ON
> > > >
> > > > (`VendorJobs`.`Industry`
> > > >
> > > > > = `StaIndTypes`.`CareerIDs`)

Re: [PHP-DB] Please help

[Micah Stevens]

echo $sql." ".$sql_ext;

Read the docs: 

http://www.php.net/echo
http://www.php.net/mysql_query



On Friday 10 September 2004 01:18 pm, you wrote:
> Now that I'm not getting an invalid error message any
> longer, how can I get it to echo the sql statement ?
>
> Stuart
>
> --- Micah Stevens <[EMAIL PROTECTED]> wrote:
> > I understand your intent, but that is not really
> > what is happening. The code
> > changes I suggested did two things:
> >
> > 1) Echo's the error statement that mysql produces
> >
> > 2) Echo's the actual SQL statement that is sent to
> > the DB
> >
> > you should be looking at #2 for answers, not your
> > intended query. This
> > 'actual' query is what I'm referring to.
> >
> > -Micah
> >
> > On Friday 10 September 2004 10:46 am, Stuart
> >
> > Felenstein wrote:
> > > I think maybe there is an implied where in the
> > > generated code, but not in my statement.  I'm
> >
> > saying
> >
> > > when I chose "ACCFIN" (as in the referred error
> > > message), the join should be looking at the
> >
> > referred
> >
> > > table , value ACCFIN .  e.g. Code_Table.CodeID
> > > (ACCFIN) = MAIN_Table.CodeID (ACCFIN) return
> >
> > label.
> >
> > > Here is the code, and I honestly am not trying to
> > > argue , just understand. Apologies if I'm
> >
> > belabouring
> >
> > > the point.
> > >
> > >  > > require('LFW3_connection.php');
> > > require('qs_functions.php');
> > > @session_start();
> > > $err_string = "";
> > > $strkeyword = "";
> > > $sql = "";
> > > $sql_ext = "";
> > > $fields = array();
> > > $fields[0] = "JobID";
> > > $fields[1] = "CompanyName";
> > > $fields[2] = "CareerCategories";
> > > $fields[3] = "City";
> > > $fields[4] = "States";
> > > $fields[5] = "AreaCode";
> > > $fields[6] = "TaxTerm";
> > > $fields[7] = "PayRate";
> > > $fields[8] = "TravelReq";
> > > $fields[9] = "Details";
> > > $fields[10] = "PostStart";
> > > $fields[11] = "JobTitle";
> > > $arryitemvalue = array();
> > > $arryitemvalue[0] = "";
> > > $arryitemvalue[1] = "";
> > > $arryitemvalue[2] = "";
> > > $arryitemvalue[3] = "";
> > > $arryitemvalue[4] = "";
> > > $arryitemvalue[5] = "";
> > > $arryitemvalue[6] = "";
> > > $arryitemvalue[7] = "";
> > > $arryitemvalue[8] = "";
> > > $arryitemvalue[9] = "";
> > > $arryitemvalue[10] = "";
> > > $arryitemvalue[11] = "";
> > > $arryopt = array();
> > > $arryopt[0] = "";
> > > $arryopt[1] = "";
> > > $arryopt[2] = "";
> > > $arryopt[3] = "";
> > > $arryopt[4] = "";
> > > $arryopt[5] = "";
> > > $arryopt[6] = "";
> > > $arryopt[7] = "";
> > > $arryopt[8] = "";
> > > $arryopt[9] = "";
> > > $arryopt[10] = "";
> > > $arryopt[11] = "";
> > > $sql .= " SELECT ";
> > > $sql .= "   VendorJobs.JobID,";
> > > $sql .= "   VendorSignUp.CompanyName,";
> > > $sql .= "   StaIndTypes.CareerCategories,";
> > > $sql .= "   StaUSCities.City,";
> > > $sql .= "   USStates.States,";
> > > $sql .= "   VendorJobs.AreaCode,";
> > > $sql .= "   staTaxTerm.TaxTerm,";
> > > $sql .= "   VendorJobs.PayRate,";
> > > $sql .= "   staTravelReq.TravelReq,";
> > > $sql .= "   VendorJobDetails.Details,";
> > > $sql .= "   VendorJobs.PostStart,";
> > > $sql .= "   VendorJobs.JobTitle";
> > > $sql .= " FROM";
> > > $sql .= "   VendorJobs";
> > > $sql .= "   INNER JOIN VendorSignUp ON
> > > (VendorJobs.VendorID = VendorSignUp.VendorID)";
> > > $sql .= "   INNER JOIN StaIndTypes ON
> > > (VendorJobs.Industry = StaIndTypes.CareerIDs)";
> > > $sql .= "   LEFT OUTER JOIN StaUSCities ON
> > > (VendorJobs.LocationCity = StaUSCities.CityID)";
> > > $sql .= "   LEFT OUTER JOIN USStates ON
> > > (VendorJobs.LocationState = USStates.StateID)";
> > > $sql .= "   LEFT OUTER JOIN staTaxTerm ON
> > > (VendorJobs.TaxTerm = staTaxTerm.TaxTermID)";
> > > $sql .= "   INNER JOIN staTravelReq ON
> > > (VendorJobs.TravelReq =
> >
> > staTravelReq.TravelReqID)";
> >
> > > $sql .= "   INNER JOIN VendorJobDetails ON
> > > (VendorJobs.JobID = VendorJobDetails.JobID)";
> > >
> > >
> > > $result = mysql_query($sql . " " . $sql_ext . "
> >
> > limit
> >
> > > 0,1")  or
> > > die(mysql_error()."".$sql." ".$sql_ext . "
> >
> > limit
> >
> > > 0,1");
> > > if (isset($_POST["QS_Submit"])) {
> > > $filter_string = "";
> > > $qry_string = "";
> > > $i = 0;
> > > while ($i < mysql_num_fields($result)) {
> > > $meta = mysql_fetch_field($result);
> > > $field_name = $meta->name;
> > > $field_type = $meta->type;
> > > if ((qsrequest("search_fd" . $i) != "") &&
> > > (qsrequest("search_fd" . $i) != "*")) {
> > > $idata = qsrequest("search_fd" . $i);
> > > if (strlen($idata) > 1) {
> > > if ($idata[strlen($idata) - 1] ==
> >
> > "*")
> >
> > > {
> > > $idata = substr($idata, 0,
> > > strlen($idata) - 1);
> > > }
> > > }
> > > $idata = str_replace("*", "%",
> >
> > $idata);
> >
> > > $irealdata = $idata;
> > > if (qsrequest("search_optfd".$i) !=
> >
> > "") {
> >
> > > 

Re: [PHP-DB] Please help

[Stuart Felenstein]

Now that I'm not getting an invalid error message any
longer, how can I get it to echo the sql statement ?

Stuart
--- Micah Stevens <[EMAIL PROTECTED]> wrote:

> 
> I understand your intent, but that is not really
> what is happening. The code 
> changes I suggested did two things: 
> 
> 1) Echo's the error statement that mysql produces
> 
> 2) Echo's the actual SQL statement that is sent to
> the DB
> 
> you should be looking at #2 for answers, not your
> intended query. This 
> 'actual' query is what I'm referring to. 
> 
> -Micah 
> 
> On Friday 10 September 2004 10:46 am, Stuart
> Felenstein wrote:
> > I think maybe there is an implied where in the
> > generated code, but not in my statement.  I'm
> saying
> > when I chose "ACCFIN" (as in the referred error
> > message), the join should be looking at the
> referred
> > table , value ACCFIN .  e.g. Code_Table.CodeID
> > (ACCFIN) = MAIN_Table.CodeID (ACCFIN) return
> label.
> >
> > Here is the code, and I honestly am not trying to
> > argue , just understand. Apologies if I'm
> belabouring
> > the point.
> >
> >  > require('LFW3_connection.php');
> > require('qs_functions.php');
> > @session_start();
> > $err_string = "";
> > $strkeyword = "";
> > $sql = "";
> > $sql_ext = "";
> > $fields = array();
> > $fields[0] = "JobID";
> > $fields[1] = "CompanyName";
> > $fields[2] = "CareerCategories";
> > $fields[3] = "City";
> > $fields[4] = "States";
> > $fields[5] = "AreaCode";
> > $fields[6] = "TaxTerm";
> > $fields[7] = "PayRate";
> > $fields[8] = "TravelReq";
> > $fields[9] = "Details";
> > $fields[10] = "PostStart";
> > $fields[11] = "JobTitle";
> > $arryitemvalue = array();
> > $arryitemvalue[0] = "";
> > $arryitemvalue[1] = "";
> > $arryitemvalue[2] = "";
> > $arryitemvalue[3] = "";
> > $arryitemvalue[4] = "";
> > $arryitemvalue[5] = "";
> > $arryitemvalue[6] = "";
> > $arryitemvalue[7] = "";
> > $arryitemvalue[8] = "";
> > $arryitemvalue[9] = "";
> > $arryitemvalue[10] = "";
> > $arryitemvalue[11] = "";
> > $arryopt = array();
> > $arryopt[0] = "";
> > $arryopt[1] = "";
> > $arryopt[2] = "";
> > $arryopt[3] = "";
> > $arryopt[4] = "";
> > $arryopt[5] = "";
> > $arryopt[6] = "";
> > $arryopt[7] = "";
> > $arryopt[8] = "";
> > $arryopt[9] = "";
> > $arryopt[10] = "";
> > $arryopt[11] = "";
> > $sql .= " SELECT ";
> > $sql .= "   VendorJobs.JobID,";
> > $sql .= "   VendorSignUp.CompanyName,";
> > $sql .= "   StaIndTypes.CareerCategories,";
> > $sql .= "   StaUSCities.City,";
> > $sql .= "   USStates.States,";
> > $sql .= "   VendorJobs.AreaCode,";
> > $sql .= "   staTaxTerm.TaxTerm,";
> > $sql .= "   VendorJobs.PayRate,";
> > $sql .= "   staTravelReq.TravelReq,";
> > $sql .= "   VendorJobDetails.Details,";
> > $sql .= "   VendorJobs.PostStart,";
> > $sql .= "   VendorJobs.JobTitle";
> > $sql .= " FROM";
> > $sql .= "   VendorJobs";
> > $sql .= "   INNER JOIN VendorSignUp ON
> > (VendorJobs.VendorID = VendorSignUp.VendorID)";
> > $sql .= "   INNER JOIN StaIndTypes ON
> > (VendorJobs.Industry = StaIndTypes.CareerIDs)";
> > $sql .= "   LEFT OUTER JOIN StaUSCities ON
> > (VendorJobs.LocationCity = StaUSCities.CityID)";
> > $sql .= "   LEFT OUTER JOIN USStates ON
> > (VendorJobs.LocationState = USStates.StateID)";
> > $sql .= "   LEFT OUTER JOIN staTaxTerm ON
> > (VendorJobs.TaxTerm = staTaxTerm.TaxTermID)";
> > $sql .= "   INNER JOIN staTravelReq ON
> > (VendorJobs.TravelReq =
> staTravelReq.TravelReqID)";
> > $sql .= "   INNER JOIN VendorJobDetails ON
> > (VendorJobs.JobID = VendorJobDetails.JobID)";
> >
> >
> > $result = mysql_query($sql . " " . $sql_ext . "
> limit
> > 0,1")  or
> > die(mysql_error()."".$sql." ".$sql_ext . "
> limit
> > 0,1");
> > if (isset($_POST["QS_Submit"])) {
> > $filter_string = "";
> > $qry_string = "";
> > $i = 0;
> > while ($i < mysql_num_fields($result)) {
> > $meta = mysql_fetch_field($result);
> > $field_name = $meta->name;
> > $field_type = $meta->type;
> > if ((qsrequest("search_fd" . $i) != "") &&
> > (qsrequest("search_fd" . $i) != "*")) {
> > $idata = qsrequest("search_fd" . $i);
> > if (strlen($idata) > 1) {
> > if ($idata[strlen($idata) - 1] ==
> "*")
> > {
> > $idata = substr($idata, 0,
> > strlen($idata) - 1);
> > }
> > }
> > $idata = str_replace("*", "%",
> $idata);
> > $irealdata = $idata;
> > if (qsrequest("search_optfd".$i) !=
> "") {
> >   $idata = qsrequest("search_optfd".
> $i) .
> > $idata ;
> > }
> > $iopt = substr($idata, 0, 2);
> > if (($iopt == "<=") || ($iopt ==
> "=<")) {
> > $iopt = "<=";
> > $irealdata = substr($idata, 2);
> > } elseif (($iopt == ">=") || ($iopt ==
> > "=>")) {
> > $iopt = ">=";
> > $irealdata = substr($idata, 2);
> > } elseif ($iopt == "==") {
> > $iopt = "=";
> >

Re: [PHP-DB] Please help

[Micah Stevens]

I understand your intent, but that is not really what is happening. The code 
changes I suggested did two things: 

1) Echo's the error statement that mysql produces

2) Echo's the actual SQL statement that is sent to the DB

you should be looking at #2 for answers, not your intended query. This 
'actual' query is what I'm referring to. 

-Micah 

On Friday 10 September 2004 10:46 am, Stuart Felenstein wrote:
> I think maybe there is an implied where in the
> generated code, but not in my statement.  I'm saying
> when I chose "ACCFIN" (as in the referred error
> message), the join should be looking at the referred
> table , value ACCFIN .  e.g. Code_Table.CodeID
> (ACCFIN) = MAIN_Table.CodeID (ACCFIN) return label.
>
> Here is the code, and I honestly am not trying to
> argue , just understand. Apologies if I'm belabouring
> the point.
>
>  require('LFW3_connection.php');
> require('qs_functions.php');
> @session_start();
> $err_string = "";
> $strkeyword = "";
> $sql = "";
> $sql_ext = "";
> $fields = array();
> $fields[0] = "JobID";
> $fields[1] = "CompanyName";
> $fields[2] = "CareerCategories";
> $fields[3] = "City";
> $fields[4] = "States";
> $fields[5] = "AreaCode";
> $fields[6] = "TaxTerm";
> $fields[7] = "PayRate";
> $fields[8] = "TravelReq";
> $fields[9] = "Details";
> $fields[10] = "PostStart";
> $fields[11] = "JobTitle";
> $arryitemvalue = array();
> $arryitemvalue[0] = "";
> $arryitemvalue[1] = "";
> $arryitemvalue[2] = "";
> $arryitemvalue[3] = "";
> $arryitemvalue[4] = "";
> $arryitemvalue[5] = "";
> $arryitemvalue[6] = "";
> $arryitemvalue[7] = "";
> $arryitemvalue[8] = "";
> $arryitemvalue[9] = "";
> $arryitemvalue[10] = "";
> $arryitemvalue[11] = "";
> $arryopt = array();
> $arryopt[0] = "";
> $arryopt[1] = "";
> $arryopt[2] = "";
> $arryopt[3] = "";
> $arryopt[4] = "";
> $arryopt[5] = "";
> $arryopt[6] = "";
> $arryopt[7] = "";
> $arryopt[8] = "";
> $arryopt[9] = "";
> $arryopt[10] = "";
> $arryopt[11] = "";
> $sql .= " SELECT ";
> $sql .= "   VendorJobs.JobID,";
> $sql .= "   VendorSignUp.CompanyName,";
> $sql .= "   StaIndTypes.CareerCategories,";
> $sql .= "   StaUSCities.City,";
> $sql .= "   USStates.States,";
> $sql .= "   VendorJobs.AreaCode,";
> $sql .= "   staTaxTerm.TaxTerm,";
> $sql .= "   VendorJobs.PayRate,";
> $sql .= "   staTravelReq.TravelReq,";
> $sql .= "   VendorJobDetails.Details,";
> $sql .= "   VendorJobs.PostStart,";
> $sql .= "   VendorJobs.JobTitle";
> $sql .= " FROM";
> $sql .= "   VendorJobs";
> $sql .= "   INNER JOIN VendorSignUp ON
> (VendorJobs.VendorID = VendorSignUp.VendorID)";
> $sql .= "   INNER JOIN StaIndTypes ON
> (VendorJobs.Industry = StaIndTypes.CareerIDs)";
> $sql .= "   LEFT OUTER JOIN StaUSCities ON
> (VendorJobs.LocationCity = StaUSCities.CityID)";
> $sql .= "   LEFT OUTER JOIN USStates ON
> (VendorJobs.LocationState = USStates.StateID)";
> $sql .= "   LEFT OUTER JOIN staTaxTerm ON
> (VendorJobs.TaxTerm = staTaxTerm.TaxTermID)";
> $sql .= "   INNER JOIN staTravelReq ON
> (VendorJobs.TravelReq = staTravelReq.TravelReqID)";
> $sql .= "   INNER JOIN VendorJobDetails ON
> (VendorJobs.JobID = VendorJobDetails.JobID)";
>
>
> $result = mysql_query($sql . " " . $sql_ext . " limit
> 0,1")  or
> die(mysql_error()."".$sql." ".$sql_ext . " limit
> 0,1");
> if (isset($_POST["QS_Submit"])) {
> $filter_string = "";
> $qry_string = "";
> $i = 0;
> while ($i < mysql_num_fields($result)) {
> $meta = mysql_fetch_field($result);
> $field_name = $meta->name;
> $field_type = $meta->type;
> if ((qsrequest("search_fd" . $i) != "") &&
> (qsrequest("search_fd" . $i) != "*")) {
> $idata = qsrequest("search_fd" . $i);
> if (strlen($idata) > 1) {
> if ($idata[strlen($idata) - 1] == "*")
> {
> $idata = substr($idata, 0,
> strlen($idata) - 1);
> }
> }
> $idata = str_replace("*", "%", $idata);
> $irealdata = $idata;
> if (qsrequest("search_optfd".$i) != "") {
>   $idata = qsrequest("search_optfd". $i) .
> $idata ;
> }
> $iopt = substr($idata, 0, 2);
> if (($iopt == "<=") || ($iopt == "=<")) {
> $iopt = "<=";
> $irealdata = substr($idata, 2);
> } elseif (($iopt == ">=") || ($iopt ==
> "=>")) {
> $iopt = ">=";
> $irealdata = substr($idata, 2);
> } elseif ($iopt == "==") {
> $iopt = "=";
> $irealdata = substr($idata, 2);
> } elseif ($iopt == "<>") {
> $irealdata = substr($idata, 2);
> } elseif ($iopt == "^^") {
>   $iopt = "*";
>   $idata =  $iopt . $irealdata . $iopt; //
> Contain
>   } elseif ($iopt == "^*") {
>   $iopt = "*";
>   $idata =  $irealdata . $iopt; // Start
> With
>   } elseif ($iopt == "*^") {
>  

Re: [PHP-DB] Please help

[Stuart Felenstein]

I think maybe there is an implied where in the
generated code, but not in my statement.  I'm saying
when I chose "ACCFIN" (as in the referred error
message), the join should be looking at the referred
table , value ACCFIN .  e.g. Code_Table.CodeID
(ACCFIN) = MAIN_Table.CodeID (ACCFIN) return label.

Here is the code, and I honestly am not trying to
argue , just understand. Apologies if I'm belabouring
the point.

".$sql." ".$sql_ext . " limit
0,1");
if (isset($_POST["QS_Submit"])) {
$filter_string = "";
$qry_string = "";
$i = 0;
while ($i < mysql_num_fields($result)) {
$meta = mysql_fetch_field($result);
$field_name = $meta->name;
$field_type = $meta->type;
if ((qsrequest("search_fd" . $i) != "") &&
(qsrequest("search_fd" . $i) != "*")) {
$idata = qsrequest("search_fd" . $i);
if (strlen($idata) > 1) {
if ($idata[strlen($idata) - 1] == "*")
{
$idata = substr($idata, 0,
strlen($idata) - 1);
}
}
$idata = str_replace("*", "%", $idata);
$irealdata = $idata;
if (qsrequest("search_optfd".$i) != "") {
  $idata = qsrequest("search_optfd". $i) .
$idata ;
}
$iopt = substr($idata, 0, 2);
if (($iopt == "<=") || ($iopt == "=<")) {
$iopt = "<=";
$irealdata = substr($idata, 2);
} elseif (($iopt == ">=") || ($iopt ==
"=>")) {
$iopt = ">=";
$irealdata = substr($idata, 2);
} elseif ($iopt == "==") {
$iopt = "=";
$irealdata = substr($idata, 2);
} elseif ($iopt == "<>") {
$irealdata = substr($idata, 2);
} elseif ($iopt == "^^") {
  $iopt = "*";
$idata =  $iopt . $irealdata . $iopt; //
Contain
  } elseif ($iopt == "^*") {
  $iopt = "*";
$idata =  $irealdata . $iopt; // Start
With
  } elseif ($iopt == "*^") {
  $iopt = "*";
$idata =  $iopt . $irealdata ; // End
With
} else {
$iopt = substr($idata, 0, 1);
if (($iopt == "<") || ($iopt == ">")
|| ($iopt == "=")) {
$irealdata = substr($idata,1);
} else {
$iopt = "=";
}
}
if (!strcasecmp($idata,"{current date and
time}")) {
$idata = time();
} elseif (!strcasecmp($idata,"{current
date}")) {
$idata = time();
} elseif (!strcasecmp($idata,"{current
time}")) {
$idata = time();
}
if ($meta) {
if ((strtolower($field_type) ==
"timestamp")
  ||(strtolower($field_type) ==
"datetime")
  ||(strtolower($field_type) ==
"smalldatetime")
  ||(strtolower($field_type) ==
"date")
  ||(strtolower($field_type) ==
"time")
  ||(strtolower($field_type) ==
"year")) {
if ((($timestamp =
strtotime($irealdata)) !== -1)) {
if ($qry_string == "") {
$qry_string = "search_fd"
. $i . "=" . urlencode($idata);
$filter_string =
$field_name . " " . $iopt . " '" . $irealdata . "'";
} else {
$qry_string .=
"&search_fd" . $i . "=" . urlencode($idata);
$filter_string .= " and "
. $field_name . " " . $iopt . " '" . $irealdata . "'";
}
} else {
$err_string .=
"Error while searching " .
$field_name . ".";
$err_string .= "Description:
Invalid DateTime.";
}
} elseif (($meta->numeric) == 1) {
if (is_numeric($irealdata)) {
if ($qry_string == "") {
$qry_string = "search_fd"
. $i . "=" . $idata;
$filter_string =
$field_name . " " . $iopt . " " . $irealdata;
} else {
$qry_string .=
"&search_fd" . $i . "=" . $idata;
$filter_string .= " and "
. $field_name . " " . $iopt . " " . $irealdata;
}
} else {
$err_string .=
"Error while searching " .
$field_name . ".";
$err_string .= "Description:
Type mismatch.";
}
} elseif ((strtolower($field_type) ==
"blob")
||(strtolower($field_type) ==
"mediumblob")
||(strtolower($field_type) ==
"longblob")) {
if ($qry_string == "") {
$qry_string = "search_fd" . $i
. "=" . urlencode(stripslashes($idata))

Fwd: Re: [PHP-DB] Please help

[Micah Stevens]

Forgot to CC the list.. 
--- Begin Message ---

there is a WHERE on the last line of the statement you sent me.

Where's are in the form of:

WHERE   

Yours is in the form:
WHERE  

You're not providing anything to compare the latter value to. You can imply 
conditions in the join syntax for sure, but the fact of the matter remains 
you are in addition specifying a syntactically incorrect WHERE statement at 
the end. 

-Micah 

On Friday 10 September 2004 10:22 am, you wrote:
> Okay, I was under the impression that "where" is
> implied inthe joins xx.xx = xx.xx .  Is that not the
> case ?
> As a matter of fact, there isn't even a where or like
> in my query.  Then again myabe that's why I can see
> *all* records but not run a search.
>
> Can you please provide an exmample of the column name
> bteween where and like ?
>
> Thank you.
> Stuart
>
> --- Micah Stevens <[EMAIL PROTECTED]> wrote:
> > If you look at the query, there's no column name
> > between 'WHERE' and 'LIKE'
> > which is a syntax error. That's the problem.
> >
> >
> > Looks like your sql generator has some issues, or
> > you didn't specify the WHERE
> > column properly. I'm not familiar with the system
> > you're using, but keep in
> > mind, that mysql_error will usually not steer you
> > wrong.
> >
> > -Micah
> >
> > On Friday 10 September 2004 10:03 am, Stuart
> >
> > Felenstein wrote:
> > > I had not, my apologies.  I think your post
> >
> > slipped
> >
> > > by.  Anyway, yes I have now inserted mysql_error()
> >
> > and
> >
> > > got my return.  Though I'm not entirely sure how
> >
> > to
> >
> > > fix it.
> > > Here is the error.
> > >
> > > You have an error in your SQL syntax. Check the
> >
> > manual
> >
> > > that corresponds to your MySQL server version for
> >
> > the
> >
> > > right syntax to use near 'Like 'ACCFIN'' at line 1
> > > SELECT `VendorJobs`.`JobID`,
> > > `VendorSignUp`.`CompanyName`,
> > > `StaIndTypes`.`CareerCategories`,
> > > `StaUSCities`.`City`, `USStates`.`States`,
> > > `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`,
> > > `VendorJobs`.`PayRate`,
> >
> > `staTravelReq`.`TravelReq`,
> >
> > > `VendorJobDetails`.`Details`,
> > > `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle`
> >
> > FROM
> >
> > > `VendorJobs` INNER JOIN `VendorSignUp` ON
> > > (`VendorJobs`.`VendorID` =
> >
> > `VendorSignUp`.`VendorID`)
> >
> > > INNER JOIN `StaIndTypes` ON
> >
> > (`VendorJobs`.`Industry` =
> >
> > > `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN
> > > `StaUSCities` ON (`VendorJobs`.`LocationCity` =
> > > `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates`
> >
> > ON
> >
> > > (`VendorJobs`.`LocationState` =
> >
> > `USStates`.`StateID`)
> >
> > > LEFT OUTER JOIN `staTaxTerm` ON
> > > (`VendorJobs`.`TaxTerm` =
> >
> > `staTaxTerm`.`TaxTermID`)
> >
> > > INNER JOIN `staTravelReq` ON
> >
> > (`VendorJobs`.`TravelReq`
> >
> > > = `staTravelReq`.`TravelReqID`) INNER JOIN
> > > `VendorJobDetails` ON (`VendorJobs`.`JobID` =
> > > `VendorJobDetails`.`JobID`) where Like 'ACCFIN'
> >
> > limit
> >
> > > 0,1
> > >
> > > --
> > >
> > > --- Micah Stevens <[EMAIL PROTECTED]>
> >
> > wrote:
> > > > did you make the change to the code I suggested?
> > > > What does MySQL say the error
> > > > is?
> > > >
> > > > -Micah
> > > >
> > > >
> > > > On Friday 10 September 2004 07:49 am, Stuart
> > > >
> > > > Felenstein wrote:
> > > > > As I said this is a code generator
> >
> > (dbqwiksite).
> >
> > > > So,
> > > >
> > > > > describing the process for creating the code
> >
> > is
> >
> > > > > different.  The $sql is fine, as far as typos
> >
> > or
> >
> > > > > incorrect characterrs.  I've gone through
> >
> > those
> >
> > > > > statement very carefully.
> > > > > I've also tried to run a debug with no luck.
> > > > > But I do know that there is something wrong
> >
> > with
> >
> > > > the
> > > >
> > > > > statement, since I can get a complete display
> >
> > of
> >
> > > > all
> > > >
> > > > > records, but can't search using criteria.
> > > > >
> > > > > Below is the $sql, one I know works , though
> >
> > I've
> >
> > > > > tried others.  VendorJobs is the table I'm
> > > >
> > > > querying ,
> > > >
> > > > > and since it's made up of values from other
> >
> > static
> >
> > > > > type tables I have all the joins in place.
> > > > >
> > > > > I've also taken the tick marks out, thinking
> >
> > maybe
> >
> > > > it
> > > >
> > > > > was a style issue.  I'm not sure where to go
> >
> > with
> >
> > > > it.
> > > >
> > > > > Getting a response from the company is like
> >
> > pretty
> >
> > > > > difficult.
> > > > >
> > > > >   `VendorJobs`.`JobID`,
> > > > >   `VendorSignUp`.`CompanyName`,
> > > > >   `StaIndTypes`.`CareerCategories`,
> > > > >   `StaUSCities`.`City`,
> > > > >   `USStates`.`States`,
> > > > >   `VendorJobs`.`AreaCode`,
> > > > >   `staTaxTerm`.`TaxTerm`,
> > > > >   `VendorJobs`.`PayRate`,
> > > > >   `staTravelReq`.`TravelReq`,
> > > > >   `VendorJobDetails`.`Details`,
> > >

Re: [PHP-DB] Please help

[Stuart Felenstein]

Okay, I was under the impression that "where" is
implied inthe joins xx.xx = xx.xx .  Is that not the
case ?
As a matter of fact, there isn't even a where or like
in my query.  Then again myabe that's why I can see
*all* records but not run a search.  

Can you please provide an exmample of the column name
bteween where and like ?

Thank you.
Stuart


--- Micah Stevens <[EMAIL PROTECTED]> wrote:

> 
> If you look at the query, there's no column name
> between 'WHERE' and 'LIKE' 
> which is a syntax error. That's the problem.
> 
> 
> Looks like your sql generator has some issues, or
> you didn't specify the WHERE 
> column properly. I'm not familiar with the system
> you're using, but keep in 
> mind, that mysql_error will usually not steer you
> wrong. 
> 
> -Micah 
> 
> On Friday 10 September 2004 10:03 am, Stuart
> Felenstein wrote:
> > I had not, my apologies.  I think your post
> slipped
> > by.  Anyway, yes I have now inserted mysql_error()
> and
> > got my return.  Though I'm not entirely sure how
> to
> > fix it.
> > Here is the error.
> >
> > You have an error in your SQL syntax. Check the
> manual
> > that corresponds to your MySQL server version for
> the
> > right syntax to use near 'Like 'ACCFIN'' at line 1
> > SELECT `VendorJobs`.`JobID`,
> > `VendorSignUp`.`CompanyName`,
> > `StaIndTypes`.`CareerCategories`,
> > `StaUSCities`.`City`, `USStates`.`States`,
> > `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`,
> > `VendorJobs`.`PayRate`,
> `staTravelReq`.`TravelReq`,
> > `VendorJobDetails`.`Details`,
> > `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle`
> FROM
> > `VendorJobs` INNER JOIN `VendorSignUp` ON
> > (`VendorJobs`.`VendorID` =
> `VendorSignUp`.`VendorID`)
> > INNER JOIN `StaIndTypes` ON
> (`VendorJobs`.`Industry` =
> > `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN
> > `StaUSCities` ON (`VendorJobs`.`LocationCity` =
> > `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates`
> ON
> > (`VendorJobs`.`LocationState` =
> `USStates`.`StateID`)
> > LEFT OUTER JOIN `staTaxTerm` ON
> > (`VendorJobs`.`TaxTerm` =
> `staTaxTerm`.`TaxTermID`)
> > INNER JOIN `staTravelReq` ON
> (`VendorJobs`.`TravelReq`
> > = `staTravelReq`.`TravelReqID`) INNER JOIN
> > `VendorJobDetails` ON (`VendorJobs`.`JobID` =
> > `VendorJobDetails`.`JobID`) where Like 'ACCFIN'
> limit
> > 0,1
> >
> > --
> >
> > --- Micah Stevens <[EMAIL PROTECTED]>
> wrote:
> > > did you make the change to the code I suggested?
> > > What does MySQL say the error
> > > is?
> > >
> > > -Micah
> > >
> > >
> > > On Friday 10 September 2004 07:49 am, Stuart
> > >
> > > Felenstein wrote:
> > > > As I said this is a code generator
> (dbqwiksite).
> > >
> > > So,
> > >
> > > > describing the process for creating the code
> is
> > > > different.  The $sql is fine, as far as typos
> or
> > > > incorrect characterrs.  I've gone through
> those
> > > > statement very carefully.
> > > > I've also tried to run a debug with no luck.
> > > > But I do know that there is something wrong
> with
> > >
> > > the
> > >
> > > > statement, since I can get a complete display
> of
> > >
> > > all
> > >
> > > > records, but can't search using criteria.
> > > >
> > > > Below is the $sql, one I know works , though
> I've
> > > > tried others.  VendorJobs is the table I'm
> > >
> > > querying ,
> > >
> > > > and since it's made up of values from other
> static
> > > > type tables I have all the joins in place.
> > > >
> > > > I've also taken the tick marks out, thinking
> maybe
> > >
> > > it
> > >
> > > > was a style issue.  I'm not sure where to go
> with
> > >
> > > it.
> > >
> > > > Getting a response from the company is like
> pretty
> > > > difficult.
> > > >
> > > >   `VendorJobs`.`JobID`,
> > > >   `VendorSignUp`.`CompanyName`,
> > > >   `StaIndTypes`.`CareerCategories`,
> > > >   `StaUSCities`.`City`,
> > > >   `USStates`.`States`,
> > > >   `VendorJobs`.`AreaCode`,
> > > >   `staTaxTerm`.`TaxTerm`,
> > > >   `VendorJobs`.`PayRate`,
> > > >   `staTravelReq`.`TravelReq`,
> > > >   `VendorJobDetails`.`Details`,
> > > >   `VendorJobs`.`PostStart`,
> > > >   `VendorJobs`.`JobTitle`
> > > > FROM
> > > >   `VendorJobs`
> > > >   INNER JOIN `VendorSignUp` ON
> > > > (`VendorJobs`.`VendorID` =
> > >
> > > `VendorSignUp`.`VendorID`)
> > >
> > > >   INNER JOIN `StaIndTypes` ON
> > >
> > > (`VendorJobs`.`Industry`
> > >
> > > > = `StaIndTypes`.`CareerIDs`)
> > > >   LEFT OUTER JOIN `StaUSCities` ON
> > > > (`VendorJobs`.`LocationCity` =
> > >
> > > `StaUSCities`.`CityID`)
> > >
> > > >   LEFT OUTER JOIN `USStates` ON
> > > > (`VendorJobs`.`LocationState` =
> > >
> > > `USStates`.`StateID`)
> > >
> > > >   LEFT OUTER JOIN `staTaxTerm` ON
> > > > (`VendorJobs`.`TaxTerm` =
> > >
> > > `staTaxTerm`.`TaxTermID`)
> > >
> > > >   INNER JOIN `staTravelReq` ON
> > > > (`VendorJobs`.`TravelReq` =
> > > > `staTravelReq`.`TravelReqID`)
> > > >   INNER JOIN `VendorJobDetails` ON
> > > > (`VendorJobs`.`JobID` =
> > >
> > > `VendorJobDetails`.`JobID`)

Re: [PHP-DB] Please help

[Micah Stevens]

If you look at the query, there's no column name between 'WHERE' and 'LIKE' 
which is a syntax error. That's the problem.


Looks like your sql generator has some issues, or you didn't specify the WHERE 
column properly. I'm not familiar with the system you're using, but keep in 
mind, that mysql_error will usually not steer you wrong. 

-Micah 

On Friday 10 September 2004 10:03 am, Stuart Felenstein wrote:
> I had not, my apologies.  I think your post slipped
> by.  Anyway, yes I have now inserted mysql_error() and
> got my return.  Though I'm not entirely sure how to
> fix it.
> Here is the error.
>
> You have an error in your SQL syntax. Check the manual
> that corresponds to your MySQL server version for the
> right syntax to use near 'Like 'ACCFIN'' at line 1
> SELECT `VendorJobs`.`JobID`,
> `VendorSignUp`.`CompanyName`,
> `StaIndTypes`.`CareerCategories`,
> `StaUSCities`.`City`, `USStates`.`States`,
> `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`,
> `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`,
> `VendorJobDetails`.`Details`,
> `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM
> `VendorJobs` INNER JOIN `VendorSignUp` ON
> (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`)
> INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` =
> `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN
> `StaUSCities` ON (`VendorJobs`.`LocationCity` =
> `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON
> (`VendorJobs`.`LocationState` = `USStates`.`StateID`)
> LEFT OUTER JOIN `staTaxTerm` ON
> (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`)
> INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq`
> = `staTravelReq`.`TravelReqID`) INNER JOIN
> `VendorJobDetails` ON (`VendorJobs`.`JobID` =
> `VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit
> 0,1
>
> --
>
> --- Micah Stevens <[EMAIL PROTECTED]> wrote:
> > did you make the change to the code I suggested?
> > What does MySQL say the error
> > is?
> >
> > -Micah
> >
> >
> > On Friday 10 September 2004 07:49 am, Stuart
> >
> > Felenstein wrote:
> > > As I said this is a code generator (dbqwiksite).
> >
> > So,
> >
> > > describing the process for creating the code is
> > > different.  The $sql is fine, as far as typos or
> > > incorrect characterrs.  I've gone through those
> > > statement very carefully.
> > > I've also tried to run a debug with no luck.
> > > But I do know that there is something wrong with
> >
> > the
> >
> > > statement, since I can get a complete display of
> >
> > all
> >
> > > records, but can't search using criteria.
> > >
> > > Below is the $sql, one I know works , though I've
> > > tried others.  VendorJobs is the table I'm
> >
> > querying ,
> >
> > > and since it's made up of values from other static
> > > type tables I have all the joins in place.
> > >
> > > I've also taken the tick marks out, thinking maybe
> >
> > it
> >
> > > was a style issue.  I'm not sure where to go with
> >
> > it.
> >
> > > Getting a response from the company is like pretty
> > > difficult.
> > >
> > >   `VendorJobs`.`JobID`,
> > >   `VendorSignUp`.`CompanyName`,
> > >   `StaIndTypes`.`CareerCategories`,
> > >   `StaUSCities`.`City`,
> > >   `USStates`.`States`,
> > >   `VendorJobs`.`AreaCode`,
> > >   `staTaxTerm`.`TaxTerm`,
> > >   `VendorJobs`.`PayRate`,
> > >   `staTravelReq`.`TravelReq`,
> > >   `VendorJobDetails`.`Details`,
> > >   `VendorJobs`.`PostStart`,
> > >   `VendorJobs`.`JobTitle`
> > > FROM
> > >   `VendorJobs`
> > >   INNER JOIN `VendorSignUp` ON
> > > (`VendorJobs`.`VendorID` =
> >
> > `VendorSignUp`.`VendorID`)
> >
> > >   INNER JOIN `StaIndTypes` ON
> >
> > (`VendorJobs`.`Industry`
> >
> > > = `StaIndTypes`.`CareerIDs`)
> > >   LEFT OUTER JOIN `StaUSCities` ON
> > > (`VendorJobs`.`LocationCity` =
> >
> > `StaUSCities`.`CityID`)
> >
> > >   LEFT OUTER JOIN `USStates` ON
> > > (`VendorJobs`.`LocationState` =
> >
> > `USStates`.`StateID`)
> >
> > >   LEFT OUTER JOIN `staTaxTerm` ON
> > > (`VendorJobs`.`TaxTerm` =
> >
> > `staTaxTerm`.`TaxTermID`)
> >
> > >   INNER JOIN `staTravelReq` ON
> > > (`VendorJobs`.`TravelReq` =
> > > `staTravelReq`.`TravelReqID`)
> > >   INNER JOIN `VendorJobDetails` ON
> > > (`VendorJobs`.`JobID` =
> >
> > `VendorJobDetails`.`JobID`)
> >
> > > Stuart
> > >
> > > --- Philip Thompson <[EMAIL PROTECTED]> wrote:
> > > > I think everyone knows that $sql is a statement.
> >
> > But
> >
> > > > what people are
> > > > asking is: what is that statement?! Because if
> >
> > there
> >
> > > > are "incorrect"
> > > > characters or 's in that statement, then that
> >
> > can
> >
> > > > break your
> > > > code/statement.
> > > >
> > > > ~Philip
> > > >
> > > > On Sep 9, 2004, at 5:49 PM, Stuart Felenstein
> >
> > wrote:
> > > > > Just getting back to this thing.  There are 3
> > > >
> > > > files
> > > >
> > > > > involved, search.php, connections.php and
> > > > > functions.php.
> > > > >
> > > > > I searched through all and couldn't find the
> > > >
> > > > meaning
> > > >

Re: [PHP-DB] Please help

[Stuart Felenstein]

I had not, my apologies.  I think your post slipped
by.  Anyway, yes I have now inserted mysql_error() and
got my return.  Though I'm not entirely sure how to
fix it.
Here is the error.

You have an error in your SQL syntax. Check the manual
that corresponds to your MySQL server version for the
right syntax to use near 'Like 'ACCFIN'' at line 1
SELECT `VendorJobs`.`JobID`,
`VendorSignUp`.`CompanyName`,
`StaIndTypes`.`CareerCategories`,
`StaUSCities`.`City`, `USStates`.`States`,
`VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`,
`VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`,
`VendorJobDetails`.`Details`,
`VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM
`VendorJobs` INNER JOIN `VendorSignUp` ON
(`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`)
INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` =
`StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN
`StaUSCities` ON (`VendorJobs`.`LocationCity` =
`StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON
(`VendorJobs`.`LocationState` = `USStates`.`StateID`)
LEFT OUTER JOIN `staTaxTerm` ON
(`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`)
INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq`
= `staTravelReq`.`TravelReqID`) INNER JOIN
`VendorJobDetails` ON (`VendorJobs`.`JobID` =
`VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit
0,1

--
--- Micah Stevens <[EMAIL PROTECTED]> wrote:

> did you make the change to the code I suggested?
> What does MySQL say the error 
> is? 
> 
> -Micah 
> 
> 
> On Friday 10 September 2004 07:49 am, Stuart
> Felenstein wrote:
> > As I said this is a code generator (dbqwiksite).
> So,
> > describing the process for creating the code is
> > different.  The $sql is fine, as far as typos or
> > incorrect characterrs.  I've gone through those
> > statement very carefully.
> > I've also tried to run a debug with no luck.
> > But I do know that there is something wrong with
> the
> > statement, since I can get a complete display of
> all
> > records, but can't search using criteria.
> >
> > Below is the $sql, one I know works , though I've
> > tried others.  VendorJobs is the table I'm
> querying ,
> > and since it's made up of values from other static
> > type tables I have all the joins in place.
> >
> > I've also taken the tick marks out, thinking maybe
> it
> > was a style issue.  I'm not sure where to go with
> it.
> > Getting a response from the company is like pretty
> > difficult.
> >
> >   `VendorJobs`.`JobID`,
> >   `VendorSignUp`.`CompanyName`,
> >   `StaIndTypes`.`CareerCategories`,
> >   `StaUSCities`.`City`,
> >   `USStates`.`States`,
> >   `VendorJobs`.`AreaCode`,
> >   `staTaxTerm`.`TaxTerm`,
> >   `VendorJobs`.`PayRate`,
> >   `staTravelReq`.`TravelReq`,
> >   `VendorJobDetails`.`Details`,
> >   `VendorJobs`.`PostStart`,
> >   `VendorJobs`.`JobTitle`
> > FROM
> >   `VendorJobs`
> >   INNER JOIN `VendorSignUp` ON
> > (`VendorJobs`.`VendorID` =
> `VendorSignUp`.`VendorID`)
> >   INNER JOIN `StaIndTypes` ON
> (`VendorJobs`.`Industry`
> > = `StaIndTypes`.`CareerIDs`)
> >   LEFT OUTER JOIN `StaUSCities` ON
> > (`VendorJobs`.`LocationCity` =
> `StaUSCities`.`CityID`)
> >   LEFT OUTER JOIN `USStates` ON
> > (`VendorJobs`.`LocationState` =
> `USStates`.`StateID`)
> >   LEFT OUTER JOIN `staTaxTerm` ON
> > (`VendorJobs`.`TaxTerm` =
> `staTaxTerm`.`TaxTermID`)
> >   INNER JOIN `staTravelReq` ON
> > (`VendorJobs`.`TravelReq` =
> > `staTravelReq`.`TravelReqID`)
> >   INNER JOIN `VendorJobDetails` ON
> > (`VendorJobs`.`JobID` =
> `VendorJobDetails`.`JobID`)
> >
> > Stuart
> >
> > --- Philip Thompson <[EMAIL PROTECTED]> wrote:
> > > I think everyone knows that $sql is a statement.
> But
> > > what people are
> > > asking is: what is that statement?! Because if
> there
> > > are "incorrect"
> > > characters or 's in that statement, then that
> can
> > > break your
> > > code/statement.
> > >
> > > ~Philip
> > >
> > > On Sep 9, 2004, at 5:49 PM, Stuart Felenstein
> wrote:
> > > > Just getting back to this thing.  There are 3
> > >
> > > files
> > >
> > > > involved, search.php, connections.php and
> > > > functions.php.
> > > >
> > > > I searched through all and couldn't find the
> > >
> > > meaning
> > >
> > > > of the $sql_ext.   $sql is just the sql
> statement
> > >
> > > I
> > >
> > > > inserted into the code.
> > > >
> > > > Great that I can't get a response from the
> > >
> > > company.
> > >
> > > > If anyone thinks it would be useful to post
> the
> > >
> > > code I
> > >
> > > > will. That is if anyone would want to see and
> look
> > >
> > > at
> > >
> > > > it.
> > > >
> > > > The weird thing is searching on one table is
> > >
> > > great.
> > >
> > > > The list (all records) works great.  The
> problem
> > > > starts when I insert all my joins.  Now I've
> > >
> > > tested
> > >
> > > > and retested the query and I know it's fine.
> > > >
> > > > Thank you ,
> > > > Stuart
> > > >
> > > > --- Steve Davies <[EMAIL PROTECTED]>
> wrote:
> > > >> What's contained in $sql and $

Re: [PHP-DB] Please help

[Micah Stevens]

did you make the change to the code I suggested? What does MySQL say the error 
is? 

-Micah 


On Friday 10 September 2004 07:49 am, Stuart Felenstein wrote:
> As I said this is a code generator (dbqwiksite). So,
> describing the process for creating the code is
> different.  The $sql is fine, as far as typos or
> incorrect characterrs.  I've gone through those
> statement very carefully.
> I've also tried to run a debug with no luck.
> But I do know that there is something wrong with the
> statement, since I can get a complete display of all
> records, but can't search using criteria.
>
> Below is the $sql, one I know works , though I've
> tried others.  VendorJobs is the table I'm querying ,
> and since it's made up of values from other static
> type tables I have all the joins in place.
>
> I've also taken the tick marks out, thinking maybe it
> was a style issue.  I'm not sure where to go with it.
> Getting a response from the company is like pretty
> difficult.
>
>   `VendorJobs`.`JobID`,
>   `VendorSignUp`.`CompanyName`,
>   `StaIndTypes`.`CareerCategories`,
>   `StaUSCities`.`City`,
>   `USStates`.`States`,
>   `VendorJobs`.`AreaCode`,
>   `staTaxTerm`.`TaxTerm`,
>   `VendorJobs`.`PayRate`,
>   `staTravelReq`.`TravelReq`,
>   `VendorJobDetails`.`Details`,
>   `VendorJobs`.`PostStart`,
>   `VendorJobs`.`JobTitle`
> FROM
>   `VendorJobs`
>   INNER JOIN `VendorSignUp` ON
> (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`)
>   INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry`
> = `StaIndTypes`.`CareerIDs`)
>   LEFT OUTER JOIN `StaUSCities` ON
> (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`)
>   LEFT OUTER JOIN `USStates` ON
> (`VendorJobs`.`LocationState` = `USStates`.`StateID`)
>   LEFT OUTER JOIN `staTaxTerm` ON
> (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`)
>   INNER JOIN `staTravelReq` ON
> (`VendorJobs`.`TravelReq` =
> `staTravelReq`.`TravelReqID`)
>   INNER JOIN `VendorJobDetails` ON
> (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`)
>
> Stuart
>
> --- Philip Thompson <[EMAIL PROTECTED]> wrote:
> > I think everyone knows that $sql is a statement. But
> > what people are
> > asking is: what is that statement?! Because if there
> > are "incorrect"
> > characters or 's in that statement, then that can
> > break your
> > code/statement.
> >
> > ~Philip
> >
> > On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote:
> > > Just getting back to this thing.  There are 3
> >
> > files
> >
> > > involved, search.php, connections.php and
> > > functions.php.
> > >
> > > I searched through all and couldn't find the
> >
> > meaning
> >
> > > of the $sql_ext.   $sql is just the sql statement
> >
> > I
> >
> > > inserted into the code.
> > >
> > > Great that I can't get a response from the
> >
> > company.
> >
> > > If anyone thinks it would be useful to post the
> >
> > code I
> >
> > > will. That is if anyone would want to see and look
> >
> > at
> >
> > > it.
> > >
> > > The weird thing is searching on one table is
> >
> > great.
> >
> > > The list (all records) works great.  The problem
> > > starts when I insert all my joins.  Now I've
> >
> > tested
> >
> > > and retested the query and I know it's fine.
> > >
> > > Thank you ,
> > > Stuart
> > >
> > > --- Steve Davies <[EMAIL PROTECTED]> wrote:
> > >> What's contained in $sql and $sql_ext ???
> > >>
> > >> Stuart Felenstein wrote:
> > >>> I'm using a product called dbqwiksite pro.  PHP
> > >>> generator for PHP - MySQL
> > >>>
> > >>> The code seems to be working fine except in my
> > >>
> > >> search
> > >>
> > >>> page where I receive an "invalid query"
> > >>>
> > >>> $result = mysql_query($sql . " " . $sql_ext . "
> > >>
> > >> limit
> > >>
> > >>> 0,1")
> > >>> or die("Invalid query");
> > >>>
> > >>> This is the place I where the code is taking the
> > >>
> > >> die
> > >>
> > >>> path.
> > >>> I'm guessing something belongs between the
> > >>
> > >> quotation
> > >>
> > >>> marks , just not sure.
> > >>>
> > >>> Anyone ?
> > >>>
> > >>> Thank you
> > >>> Stuart
> > >>
> > >> --
> > >> PHP Database Mailing List (http://www.php.net/)
> > >> To unsubscribe, visit:
> >
> > http://www.php.net/unsub.php
> >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, visit:
> >
> > http://www.php.net/unsub.php
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Please help

[Stuart Felenstein]

As I said this is a code generator (dbqwiksite). So,
describing the process for creating the code is
different.  The $sql is fine, as far as typos or
incorrect characterrs.  I've gone through those
statement very carefully.
I've also tried to run a debug with no luck.
But I do know that there is something wrong with the
statement, since I can get a complete display of all
records, but can't search using criteria.

Below is the $sql, one I know works , though I've
tried others.  VendorJobs is the table I'm querying ,
and since it's made up of values from other static
type tables I have all the joins in place. 

I've also taken the tick marks out, thinking maybe it
was a style issue.  I'm not sure where to go with it. 
Getting a response from the company is like pretty
difficult.
 
  `VendorJobs`.`JobID`,
  `VendorSignUp`.`CompanyName`,
  `StaIndTypes`.`CareerCategories`,
  `StaUSCities`.`City`,
  `USStates`.`States`,
  `VendorJobs`.`AreaCode`,
  `staTaxTerm`.`TaxTerm`,
  `VendorJobs`.`PayRate`,
  `staTravelReq`.`TravelReq`,
  `VendorJobDetails`.`Details`,
  `VendorJobs`.`PostStart`,
  `VendorJobs`.`JobTitle`
FROM
  `VendorJobs`
  INNER JOIN `VendorSignUp` ON
(`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`)
  INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry`
= `StaIndTypes`.`CareerIDs`)
  LEFT OUTER JOIN `StaUSCities` ON
(`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`)
  LEFT OUTER JOIN `USStates` ON
(`VendorJobs`.`LocationState` = `USStates`.`StateID`)
  LEFT OUTER JOIN `staTaxTerm` ON
(`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`)
  INNER JOIN `staTravelReq` ON
(`VendorJobs`.`TravelReq` =
`staTravelReq`.`TravelReqID`)
  INNER JOIN `VendorJobDetails` ON
(`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`)

Stuart
--- Philip Thompson <[EMAIL PROTECTED]> wrote:

> I think everyone knows that $sql is a statement. But
> what people are 
> asking is: what is that statement?! Because if there
> are "incorrect" 
> characters or 's in that statement, then that can
> break your 
> code/statement.
> 
> ~Philip
> 
> On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote:
> 
> > Just getting back to this thing.  There are 3
> files
> > involved, search.php, connections.php and
> > functions.php.
> >
> > I searched through all and couldn't find the
> meaning
> > of the $sql_ext.   $sql is just the sql statement
> I
> > inserted into the code.
> >
> > Great that I can't get a response from the
> company.
> > If anyone thinks it would be useful to post the
> code I
> > will. That is if anyone would want to see and look
> at
> > it.
> >
> > The weird thing is searching on one table is
> great.
> > The list (all records) works great.  The problem
> > starts when I insert all my joins.  Now I've
> tested
> > and retested the query and I know it's fine.
> >
> > Thank you ,
> > Stuart
> >
> >
> > --- Steve Davies <[EMAIL PROTECTED]> wrote:
> >
> >> What's contained in $sql and $sql_ext ???
> >>
> >>
> >> Stuart Felenstein wrote:
> >>
> >>> I'm using a product called dbqwiksite pro.  PHP
> >>> generator for PHP - MySQL
> >>>
> >>> The code seems to be working fine except in my
> >> search
> >>> page where I receive an "invalid query"
> >>>
> >>> $result = mysql_query($sql . " " . $sql_ext . "
> >> limit
> >>> 0,1")
> >>> or die("Invalid query");
> >>>
> >>> This is the place I where the code is taking the
> >> die
> >>> path.
> >>> I'm guessing something belongs between the
> >> quotation
> >>> marks , just not sure.
> >>>
> >>> Anyone ?
> >>>
> >>> Thank you
> >>> Stuart
> >>>
> >>>
> >>>
> >>
> >> -- 
> >> PHP Database Mailing List (http://www.php.net/)
> >> To unsubscribe, visit:
> http://www.php.net/unsub.php
> >>
> >>
> >
> > -- 
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit:
> http://www.php.net/unsub.php
> >
> 
> -- 
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 
> 

-- 
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To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Please help

[Philip Thompson]

I think everyone knows that $sql is a statement. But what people are 
asking is: what is that statement?! Because if there are "incorrect" 
characters or 's in that statement, then that can break your 
code/statement.

~Philip
On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote:
Just getting back to this thing.  There are 3 files
involved, search.php, connections.php and
functions.php.
I searched through all and couldn't find the meaning
of the $sql_ext.   $sql is just the sql statement I
inserted into the code.
Great that I can't get a response from the company.
If anyone thinks it would be useful to post the code I
will. That is if anyone would want to see and look at
it.
The weird thing is searching on one table is great.
The list (all records) works great.  The problem
starts when I insert all my joins.  Now I've tested
and retested the query and I know it's fine.
Thank you ,
Stuart
--- Steve Davies <[EMAIL PROTECTED]> wrote:
What's contained in $sql and $sql_ext ???
Stuart Felenstein wrote:
I'm using a product called dbqwiksite pro.  PHP
generator for PHP - MySQL
The code seems to be working fine except in my
search
page where I receive an "invalid query"
$result = mysql_query($sql . " " . $sql_ext . "
limit
0,1")
or die("Invalid query");
This is the place I where the code is taking the
die
path.
I'm guessing something belongs between the
quotation
marks , just not sure.
Anyone ?
Thank you
Stuart

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Re: [PHP-DB] Please help

[Stuart Felenstein]

Just getting back to this thing.  There are 3 files
involved, search.php, connections.php and
functions.php.

I searched through all and couldn't find the meaning
of the $sql_ext.   $sql is just the sql statement I
inserted into the code.  

Great that I can't get a response from the company.  
If anyone thinks it would be useful to post the code I
will. That is if anyone would want to see and look at
it. 

The weird thing is searching on one table is great. 
The list (all records) works great.  The problem
starts when I insert all my joins.  Now I've tested
and retested the query and I know it's fine.

Thank you ,
Stuart


--- Steve Davies <[EMAIL PROTECTED]> wrote:

> What's contained in $sql and $sql_ext ???
> 
> 
> Stuart Felenstein wrote:
> 
> >I'm using a product called dbqwiksite pro.  PHP
> >generator for PHP - MySQL 
> >
> >The code seems to be working fine except in my
> search
> >page where I receive an "invalid query"
> >
> >$result = mysql_query($sql . " " . $sql_ext . "
> limit
> >0,1")
> >or die("Invalid query");
> >
> >This is the place I where the code is taking the
> die
> >path.
> >I'm guessing something belongs between the
> quotation
> >marks , just not sure.
> >
> >Anyone ?
> >
> >Thank you
> >Stuart
> >
> >  
> >
> 
> -- 
> PHP Database Mailing List (http://www.php.net/)
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> 
> 

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Re: [PHP-DB] Please help

[Steve Davies]

What's contained in $sql and $sql_ext ???
Stuart Felenstein wrote:
I'm using a product called dbqwiksite pro.  PHP
generator for PHP - MySQL 

The code seems to be working fine except in my search
page where I receive an "invalid query"
$result = mysql_query($sql . " " . $sql_ext . " limit
0,1")
or die("Invalid query");
This is the place I where the code is taking the die
path.
I'm guessing something belongs between the quotation
marks , just not sure.
Anyone ?
Thank you
Stuart
 

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RE: [PHP-DB] Please help

[Gary Every]

What exactly is in the $sql and $sql_ext??

Try to put it into a variable so you can print out what you're sending
to the DB

Such as:
$query = $sql . " " . $sql_ext . " limit 0,1";
Echo $query . "";

If your $sql and $sql_ext combine to make a valid sql query, the " " is
just putting a space inside the query, and it is valid

Gary Every
Sr. UNIX Administrator
Ingram Entertainment Inc.
2 Ingram Blvd, La Vergne, TN 37089
"Pay It Forward!"


-Original Message-
From: Stuart Felenstein [mailto:[EMAIL PROTECTED] 
Sent: Wednesday, September 08, 2004 2:29 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Please help


I'm using a product called dbqwiksite pro.  PHP
generator for PHP - MySQL 

The code seems to be working fine except in my search
page where I receive an "invalid query"

$result = mysql_query($sql . " " . $sql_ext . " limit
0,1")
or die("Invalid query");

This is the place I where the code is taking the die
path.
I'm guessing something belongs between the quotation
marks , just not sure.

Anyone ?

Thank you
Stuart

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Re: [PHP-DB] Please help

[Micah Stevens]

Change the code to this:

$result = mysql_query($sql . " " . $sql_ext . " limit 0,1")  or 
die(mysql_error()."".$sql." ".$sql_ext . " limit 0,1");

And you'll get the error from mysql and a copy of the actual query. Without 
that, it's pretty tough to help. 

-Micah 

On Wednesday 08 September 2004 12:29 pm, Stuart Felenstein wrote:
> I'm using a product called dbqwiksite pro.  PHP
> generator for PHP - MySQL
>
> The code seems to be working fine except in my search
> page where I receive an "invalid query"
>
> $result = mysql_query($sql . " " . $sql_ext . " limit
> 0,1")
> or die("Invalid query");
>
> This is the place I where the code is taking the die
> path.
> I'm guessing something belongs between the quotation
> marks , just not sure.
>
> Anyone ?
>
> Thank you
> Stuart

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Re: [PHP-DB] Please help

[Stuart Felenstein]

Not sure what the varialbe are set too.  Going through
the code and rebuilding the query.
Post on hold to further notice :).

Stuart
--- Peter Ellis <[EMAIL PROTECTED]> wrote:

> What are the variables $sql and $sql_ext set to? 
> The query format is
> fine.  It's probably in your variables.
> -- 
> Peter Ellis - [EMAIL PROTECTED]
> Web Design and Development Consultant
> naturalaxis | http://www.naturalaxis.com/
> 
> On Wed, 2004-09-08 at 12:29 -0700, Stuart Felenstein
> wrote:
> > I'm using a product called dbqwiksite pro.  PHP
> > generator for PHP - MySQL 
> > 
> > The code seems to be working fine except in my
> search
> > page where I receive an "invalid query"
> > 
> > $result = mysql_query($sql . " " . $sql_ext . "
> limit
> > 0,1")
> > or die("Invalid query");
> > 
> > This is the place I where the code is taking the
> die
> > path.
> > I'm guessing something belongs between the
> quotation
> > marks , just not sure.
> > 
> > Anyone ?
> > 
> > Thank you
> > Stuart
> 
> 
> 

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RE: [PHP-DB] Please help

[Peter Lovatt]

HI

try 

 $result = mysql_query($sql . " " . $sql_ext . " limit 0,1")
 or die("Invalid query".mysql_error());

which will give more detail on the error.

HTH

Peter

> -Original Message-
> From: Stuart Felenstein [mailto:[EMAIL PROTECTED]
> Sent: 08 September 2004 20:29
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Please help
> 
> 
> I'm using a product called dbqwiksite pro.  PHP
> generator for PHP - MySQL 
> 
> The code seems to be working fine except in my search
> page where I receive an "invalid query"
> 
> $result = mysql_query($sql . " " . $sql_ext . " limit
> 0,1")
> or die("Invalid query");
> 
> This is the place I where the code is taking the die
> path.
> I'm guessing something belongs between the quotation
> marks , just not sure.
> 
> Anyone ?
> 
> Thank you
> Stuart
> 
> -- 
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 

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Re: [PHP-DB] Please help

[Peter Ellis]

What are the variables $sql and $sql_ext set to?  The query format is
fine.  It's probably in your variables.
-- 
Peter Ellis - [EMAIL PROTECTED]
Web Design and Development Consultant
naturalaxis | http://www.naturalaxis.com/

On Wed, 2004-09-08 at 12:29 -0700, Stuart Felenstein wrote:
> I'm using a product called dbqwiksite pro.  PHP
> generator for PHP - MySQL 
> 
> The code seems to be working fine except in my search
> page where I receive an "invalid query"
> 
> $result = mysql_query($sql . " " . $sql_ext . " limit
> 0,1")
> or die("Invalid query");
> 
> This is the place I where the code is taking the die
> path.
> I'm guessing something belongs between the quotation
> marks , just not sure.
> 
> Anyone ?
> 
> Thank you
> Stuart

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[PHP-DB] Please help

[Stuart Felenstein]

I'm using a product called dbqwiksite pro.  PHP
generator for PHP - MySQL 

The code seems to be working fine except in my search
page where I receive an "invalid query"

$result = mysql_query($sql . " " . $sql_ext . " limit
0,1")
or die("Invalid query");

This is the place I where the code is taking the die
path.
I'm guessing something belongs between the quotation
marks , just not sure.

Anyone ?

Thank you
Stuart

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RE: [PHP-DB] Please help - Managing relational tables using PHP

[Michael Knauf/Niles]

Well, you can use the "SQL" tab to write a sql statement to do whatever you
want, but no, the graphical interface doesn't appear to do that.



   

  "Phanivas Vemuri"

  <[EMAIL PROTECTED]To:   [EMAIL PROTECTED], 
[EMAIL PROTECTED]
  m>   cc: 

               Subject:  RE: [PHP-DB] Please help - 
Managing relational tables using PHP   
  01/02/03 04:34 PM

   

   





Dear Mr. Ryan,
May be I did not present my question clearly to you. My question in detail
is ...
--> Does phpMyAdmin supports / allow / interface, editing data from "one to

many" or ,"many to many" related tables, using a single phpMyAdmin
interface( a single HTML page)

I have a (primary) table and one of its fields(category) has one or many
values , all these values are also listed in another(secondary) table. I
would like to know if phpMyAdmin provides any interface such that those
values from the secondary table appears(listed) (for example : as a Form)
in
the field of the primary table.

PrimaryTable(X,Y,category,Z) pk:X
SecondaryTable(category)pk:category
I am not intending to enforce referential integrity. My question is just
about the interfacing.
Thank you,
Phani.


>From: "Ryan Jameson (USA)" <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Subject: RE: [PHP-DB] Please help - Managing relational tables using PHP
>Date: Thu, 2 Jan 2003 13:32:15 -0700
>
>You need to realize that phpMyAdmin is a database manangement tool, not a
>database. So I'll answer your questions based on the mySQL database ...
>
>1. MySQL 3.23.43b introduced a table type of InnoDB which is the first to
>allow foreign key constraints. http://www.mysql.com/doc/en/SEC448.html
...
>However, you do not need to enforce constraints to have a relationship. I
>just assume that's what you're getting at.
>
>2. Yes, if you enforce constraints with the CASCADE option it will. ..see
>same document: http://www.mysql.com/doc/en/SEC448.html
>
>
>... Now the catch, my version of phpMyAdmin 2.3.0 does not allow me to
>choose the table type of InnoDB, even though my database is version
>3.23.54. So I assume phpMyAdmin does not allow one to manage these
>attributes. Someone please correct me if I'm wrong.
>
><>< Ryan
>
>
>
>-Original Message-
>From: Phanivas Vemuri [mailto:[EMAIL PROTECTED]]
>Sent: Thursday, January 02, 2003 12:50 PM
>To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED]
>Subject: [PHP-DB] Please help - Managing relational tables using PHP
>
>
>hi all,
>1. How does phpMyAdmin manage tables with "one to many" or "many to one"
>relation ships. Is there any documentation about this.
> --or-
>2. Does phpMyAdmin support editing data from more than one table that have

>a
>one to many or many to one relationship.
>
>Thanks and Wish you a happy new year,
>Phani.
>
>_
>MSN 8 with e-mail virus protection service: 2 months FREE*
>http://join.msn.com/?page=features/virus
>
>
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>
>
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This mail is confidential







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RE: [PHP-DB] Please help - Managing relational tables using PHP

[Phanivas Vemuri]

Dear Mr. Ryan,
May be I did not present my question clearly to you. My question in detail 
is ...
--> Does phpMyAdmin supports / allow / interface, editing data from "one to 
many" or ,"many to many" related tables, using a single phpMyAdmin 
interface( a single HTML page)

I have a (primary) table and one of its fields(category) has one or many 
values , all these values are also listed in another(secondary) table. I 
would like to know if phpMyAdmin provides any interface such that those 
values from the secondary table appears(listed) (for example : as a Form) in 
the field of the primary table.

PrimaryTable(X,Y,category,Z) pk:X
SecondaryTable(category)pk:category
I am not intending to enforce referential integrity. My question is just 
about the interfacing.
Thank you,
Phani.


From: "Ryan Jameson (USA)" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Subject: RE: [PHP-DB] Please help - Managing relational tables using PHP
Date: Thu, 2 Jan 2003 13:32:15 -0700

You need to realize that phpMyAdmin is a database manangement tool, not a 
database. So I'll answer your questions based on the mySQL database ...

1. MySQL 3.23.43b introduced a table type of InnoDB which is the first to 
allow foreign key constraints. http://www.mysql.com/doc/en/SEC448.html  ... 
However, you do not need to enforce constraints to have a relationship. I 
just assume that's what you're getting at.

2. Yes, if you enforce constraints with the CASCADE option it will. ..see 
same document: http://www.mysql.com/doc/en/SEC448.html


... Now the catch, my version of phpMyAdmin 2.3.0 does not allow me to 
choose the table type of InnoDB, even though my database is version 
3.23.54. So I assume phpMyAdmin does not allow one to manage these 
attributes. Someone please correct me if I'm wrong.

<>< Ryan



-Original Message-
From: Phanivas Vemuri [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 02, 2003 12:50 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: [PHP-DB] Please help - Managing relational tables using PHP


hi all,
1. How does phpMyAdmin manage tables with "one to many" or "many to one"
relation ships. Is there any documentation about this.
--or-
2. Does phpMyAdmin support editing data from more than one table that have 
a
one to many or many to one relationship.

Thanks and Wish you a happy new year,
Phani.

_
MSN 8 with e-mail virus protection service: 2 months FREE*
http://join.msn.com/?page=features/virus


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RE: [PHP-DB] Please help - Managing relational tables using PHP

[Ryan Jameson (USA)]

You need to realize that phpMyAdmin is a database manangement tool, not a database. So 
I'll answer your questions based on the mySQL database ... 

1. MySQL 3.23.43b introduced a table type of InnoDB which is the first to allow 
foreign key constraints. http://www.mysql.com/doc/en/SEC448.html  ... However, you do 
not need to enforce constraints to have a relationship. I just assume that's what 
you're getting at.

2. Yes, if you enforce constraints with the CASCADE option it will. ..see same 
document: http://www.mysql.com/doc/en/SEC448.html


... Now the catch, my version of phpMyAdmin 2.3.0 does not allow me to choose the 
table type of InnoDB, even though my database is version 3.23.54. So I assume 
phpMyAdmin does not allow one to manage these attributes. Someone please correct me if 
I'm wrong.

<>< Ryan



-Original Message-
From: Phanivas Vemuri [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 02, 2003 12:50 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: [PHP-DB] Please help - Managing relational tables using PHP


hi all,
1. How does phpMyAdmin manage tables with "one to many" or "many to one" 
relation ships. Is there any documentation about this.
--or-
2. Does phpMyAdmin support editing data from more than one table that have a 
one to many or many to one relationship.

Thanks and Wish you a happy new year,
Phani.

_
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http://join.msn.com/?page=features/virus


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[PHP-DB] Please help - Managing relational tables using PHP

[Phanivas Vemuri]

hi all,
1. How does phpMyAdmin manage tables with "one to many" or "many to one" 
relation ships. Is there any documentation about this.
   --or-
2. Does phpMyAdmin support editing data from more than one table that have a 
one to many or many to one relationship.

Thanks and Wish you a happy new year,
Phani.

_
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http://join.msn.com/?page=features/virus


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[PHP-DB] Please Help

[Petre NICOARA]

Hi, guys
I have a small and delicate problem. This I think, involves using the
rewriterule from apache server, but I don't know where to post it Here's
my situation : 
Until recently, my client had his site on a stack server. In this way,
all his pages where addressed by :
http://www.sitename.com/sitename/page.php
also, google and all other search engines indexed his pages regarding
his old location now, the problem is that, he's moved from the last
host, and now he have pages in the root folder, not on a stack,
addressed by http://www.sitename.com/pagename.extension
for the php pages, I made a redirect in a sitename folder, in order to
get, by redirect, to the new location. Now, some new facts on revealed
that google, which is his reference tool regarding web placing, google
doesn't like long addresses and also doesn't like php parameters. So,
here's my new job I have to find a way of transforming the requests (
http://www.sitename.com/sitename/page.extension?parameters ) into new
web addresses (
http://www.sitename.com/parameter1/value/parameter2/value/phppagename.ht

ml )
So, I have to modify .htaccess file, to use a RewirteRule
And, here's the content of .htaccess file 
 
ErrorDocument 404 http://www.sitename.com/index.php
 
RewriteEngine on
RewriteBase /
RewriteRule /sitename/([a-z0-9]+).htm$ /$1.php [R] ( these are for the
1st problem)
 
Some people recommended using rewrite module, and alias, but I have no
idea how this is done. When I use the code above the only think I see
when I click on a search result link is :
 
Forbidden
You don't have permission to access.../sitename/..
 
Now I have to make the parameter modification. So if the user come to
this address :
http://www.sitename.com/zone/1/country/31/city/2/location.html in fact
the server will access this page
http://www.sitename.com/location.php?zone=1

&country=31&city=2
 
Do you have any idea how this is done ? Can you give me some link, or
point me to some direction . Thank you
 
 
 
 
 



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[PHP-DB] please help.. serial number generator

[rainydays_sunshine]

Any ideas how to generate a unique serial number that you can check that
it's valid..

Thanks...



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Re: [PHP-DB] please help!

[szii]

You're not evaluating the variable correctly...
Instead of... 
$site = ".row['site'].";   
It should be...
$site = $row["site"];
Or
$site = "$row['site']"

I'm not sure why you have leading and trailing
periods, either.  Looks like you were trying to 
do string appends, but they're not being used
correctly

'Luck

-Szii

- Original Message - 
From: "Evans, Josh" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, April 23, 2002 11:26 AM
Subject: [PHP-DB] please help!


> Database: sites
> 
> I have to columns: "id" and "site". I will have approx 800 entries in the db
> when I am finished and what I want it to do and start with 1st entry and
> ping the site and display in a table on the screen if it was able to ping
> site or not.
> 
> This is what I have so far. My problem is I am not sure how to make it
> select a entry, perform the ping, record the results, and then continue to
> the next entry.
> 
>  $db = mysql_connect("localhost","","") or die("Problem connecting");
> mysql_select_db("josh") or die("Problem selecting database");
> $query = "SELECT * FROM josh.sites";
> $result = mysql_query($query) or die ("Query failed");
> $numofrows = mysql_num_rows($result);
> for($i = 0; $i < $numofrows; $i++) {
> $row = mysql_fetch_array($result);
> if($i % 2) { //this means if there is a remainder
> 
> $site = ".row['site'].";   < its not pulling a site from the
> database?
> 
> 
> $host1 = `ping -n 4 $site`;
> 
> if (eregi("reply", $host1)) {
> echo "$site is up!";
> } else {
> echo ("$site is down!");
> }
> 
> echo ("");
> ?>
> 
> Please Help!
> 
> Josh Evans
> 
> 
> 
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[PHP-DB] please help!

[Evans, Josh]

Database: sites

I have to columns: "id" and "site". I will have approx 800 entries in the db
when I am finished and what I want it to do and start with 1st entry and
ping the site and display in a table on the screen if it was able to ping
site or not.

This is what I have so far. My problem is I am not sure how to make it
select a entry, perform the ping, record the results, and then continue to
the next entry.

");
?>

Please Help!

Josh Evans



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RE: [PHP-DB] Please help count ?

[Rosser, Chris]

Dave,

Try:

$postcode_parts = explode(" ", $postcode);

The first part of the postcode will be available in $postcode_parts[0].


- Chris


> -Original Message-
> From: Dave Carrera [SMTP:[EMAIL PROTECTED]]
> Sent: Thursday, April 04, 2002 5:29 PM
> To:   [EMAIL PROTECTED]
> Subject:  [PHP-DB] Please help count ?
> 
> Hi All
> 
>  
> 
> I have a variable returned by my application.
> 
>  
> 
> For clarity it is a post code so result could be
> 
>  
> 
> EX3 T56 or BG56 G67 or CA2 123
> 
>  
> 
> But I only need the first bit of the postcode to continue my search.
> 
>  
> 
> How do I set my var to only include the first bit of the postcode.
> 
>  
> 
> It is returned in the format shown, so my question might be how I read
> only the first bit before the space.
> 
>  
> 
> I fully appreciate any kind of help with this and as always thank you in
> advance for any help.
> 
>  
> 
>  
> 
>  
> 
> Dave Carrera
> 
> Php Developer
> 
> http://davecarrera.freelancers.net
> 
> http://www.davecarrera.com
> 
>  
> 
>  
> 

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RE: [PHP-DB] Please help count ?

[Rick Emery]

$post = ereg_replace("(.*) ","\\1", $old_post);

If you want to search mysql for this that is something else

-Original Message-
From: Dave Carrera [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 04, 2002 10:29 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Please help count ?


Hi All

 

I have a variable returned by my application.

 

For clarity it is a post code so result could be

 

EX3 T56 or BG56 G67 or CA2 123

 

But I only need the first bit of the postcode to continue my search.

 

How do I set my var to only include the first bit of the postcode.

 

It is returned in the format shown, so my question might be how I read
only the first bit before the space.

 

I fully appreciate any kind of help with this and as always thank you in
advance for any help.

 

 

 

Dave Carrera

Php Developer

http://davecarrera.freelancers.net

http://www.davecarrera.com

 

 


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Re: [PHP-DB] Please help count ?

[Shane Wright]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1


you can do this...

$myvar = ecplode(' ', $postcode);
$firsthalf = $myvar[0];

But not everyone puts the space in there

S

On Thursday 04 April 2002 5:29 pm, Dave Carrera wrote:
> Hi All
>
>
>
> I have a variable returned by my application.
>
>
>
> For clarity it is a post code so result could be
>
>
>
> EX3 T56 or BG56 G67 or CA2 123
>
>
>
> But I only need the first bit of the postcode to continue my search.
>
>
>
> How do I set my var to only include the first bit of the postcode.
>
>
>
> It is returned in the format shown, so my question might be how I read
> only the first bit before the space.
>
>
>
> I fully appreciate any kind of help with this and as always thank you in
> advance for any help.
>
>
>
>
>
>
>
> Dave Carrera
>
> Php Developer
>
> http://davecarrera.freelancers.net
>
> http://www.davecarrera.com

- -- 
Shane
-BEGIN PGP SIGNATURE-
Version: GnuPG v1.0.6 (GNU/Linux)
Comment: For info see http://www.gnupg.org

iD8DBQE8rH/75DXg6dCMBrQRAlEUAJ9R/jfY0Ufdj5yCSiTRw7qB81B3TwCeJELP
GW03u2HKvFXy9fDEEdV1lfg=
=xALC
-END PGP SIGNATURE-


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[PHP-DB] Please help count ?

[Dave Carrera]

Hi All

 

I have a variable returned by my application.

 

For clarity it is a post code so result could be

 

EX3 T56 or BG56 G67 or CA2 123

 

But I only need the first bit of the postcode to continue my search.

 

How do I set my var to only include the first bit of the postcode.

 

It is returned in the format shown, so my question might be how I read
only the first bit before the space.

 

I fully appreciate any kind of help with this and as always thank you in
advance for any help.

 

 

 

Dave Carrera

Php Developer

http://davecarrera.freelancers.net

http://www.davecarrera.com

 

 

RE: [PHP-DB] Please help me! :(

[Howard Picken]

Can I ask why you're not using "autoincrement" for your
id field? If you used this you would never have the problem
your having.

Everytime a record is added it will increment the id field
by one, so you don't have to use all the code your using.

Howard

-Original Message-
From: Leif K-Brooks [mailto:[EMAIL PROTECTED]]
Sent: Sunday, 24 March 2002 3:42 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Please help me! :(


I have a site where things in the database have ids.  When something new is
added, it gets an id one higher than the highest existing id.  I use code
something like this:

$gethighestid = mysql_fetch_array(mysql_query("select id from table order by
id desc limit 1"));
$tobeid = $gethighestid[id]+1;
mysql_query("insert into table(id,othercolumn,othercolumn2)
values('$tobeid','something','something')");

The thing is, I just got two rows with duplicate ids.  Aparantley, two
people must of added two things at just the right times to make the same id.
Is there any way that will reduce, or eliminate, the time gap between
getting id and inserting?

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[PHP-DB] Please help me! :(

[Leif K-Brooks]

I have a site where things in the database have ids.  When something new is
added, it gets an id one higher than the highest existing id.  I use code
something like this:

$gethighestid = mysql_fetch_array(mysql_query("select id from table order by
id desc limit 1"));
$tobeid = $gethighestid[id]+1;
mysql_query("insert into table(id,othercolumn,othercolumn2)
values('$tobeid','something','something')");

The thing is, I just got two rows with duplicate ids.  Aparantley, two
people must of added two things at just the right times to make the same id.
Is there any way that will reduce, or eliminate, the time gap between
getting id and inserting? 

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Re: [PHP-DB] PLEASE HELP !!!!!!!

[DL Neil]

Manual reference:
http://www.php.net/manual/en/function.mysql-connect.php

resource mysql_connect ( [string server [, string username [, string
password [, bool new_link)

Seeing it is not a web page you're connecting to, try removing the
"http://";.

"localhost" has particular meaning within the IP conventions - if
cgi.xyz.com is remote then only use that server's name and remove the
'local' designation, ie use the name from your computer's perspective,
not the remote computer's perspective. Use PING to ascertain that you
have the server's name expressed correctly.

The default values are "server =  'localhost:3306' " which may be
causing some confusion here. I recommend simplifying by leaving out the
port number (3306) for now - MySQL knows that (hopefully) and thus the
remote computer's RDBMS should be 'listening' on that port - if it's
not, we'll have to get downer-and-dirtier...

Let us know how you get on!
=dn

- Original Message -
From: "Beau Lebens" <[EMAIL PROTECTED]>
To: "'Inter-Media Webmaster'" <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: 14 March 2002 00:21
Subject: RE: [PHP-DB] PLEASE HELP !!!


> try an IP
>
>
> // -Original Message-
> // From: Inter-Media Webmaster [mailto:[EMAIL PROTECTED]]
> // Sent: Thursday, 14 March 2002 6:00 AM
> // To: [EMAIL PROTECTED]
> // Subject: [PHP-DB] PLEASE HELP !!!
> //
> //
> // PLEASE HELP !!!
> // HOW I CAN GET CONNETCTION TO REMOTE HOST IN OTHER SERVER WITH
> // mysql_connect();
> // e.g.
> // my php file is in www.abc.com/index.php
> // mysql server is localhost on cgi.xyz.com
> //
> // can i do something like this
> //
> // mysql_connect("http://cgi.xyz.com:localhost:3306";, username,
> // password);
> //
> // PLEASE ANSWER to [EMAIL PROTECTED]
> //
> //
> //
> //
> //
> // --
> // PHP Database Mailing List (http://www.php.net/)
> // To unsubscribe, visit: http://www.php.net/unsub.php
> //
>
> --
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RE: [PHP-DB] PLEASE HELP !!!!!!!

[Beau Lebens]

try an IP


// -Original Message-
// From: Inter-Media Webmaster [mailto:[EMAIL PROTECTED]]
// Sent: Thursday, 14 March 2002 6:00 AM
// To: [EMAIL PROTECTED]
// Subject: [PHP-DB] PLEASE HELP !!!
// 
// 
// PLEASE HELP !!!
// HOW I CAN GET CONNETCTION TO REMOTE HOST IN OTHER SERVER WITH
// mysql_connect();
// e.g.
// my php file is in www.abc.com/index.php
// mysql server is localhost on cgi.xyz.com
// 
// can i do something like this
// 
// mysql_connect("http://cgi.xyz.com:localhost:3306";, username, 
// password);
// 
// PLEASE ANSWER to [EMAIL PROTECTED]
// 
// 
// 
// 
// 
// -- 
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[PHP-DB] PLEASE HELP !!!!!!!

[Inter-Media Webmaster]

PLEASE HELP !!!
HOW I CAN GET CONNETCTION TO REMOTE HOST IN OTHER SERVER WITH
mysql_connect();
e.g.
my php file is in www.abc.com/index.php
mysql server is localhost on cgi.xyz.com

can i do something like this

mysql_connect("http://cgi.xyz.com:localhost:3306";, username, password);

PLEASE ANSWER to [EMAIL PROTECTED]





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[PHP-DB] PLEASE HELP mysql_connect()!!!

[Inter-Media Webmaster]

PLEASE HELP !!!
HOW I CAN GET CONNETCTION TO REMOTE HOST IN OTHER SERVER WITH
mysql_connect();
e.g.
my php file is in www.abc.com/index.php
mysql server is in cgi.xyz.com on localhost

PLEASE ANSWER to [EMAIL PROTECTED]



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Re: [PHP-DB] Please help, On phpnuke and phpgroupware.

[Jason Wong]

On Thursday 10 January 2002 22:15, HiM wrote:
> Jason,
> phpNuke problem fixed after downgrade to 4.0.6

Glad to hear.

> Thanks !

You're welcome :)



-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
Falling in love is a lot like dying.  You never get to do it enough to
become good at it.
*/

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Re: [PHP-DB] Please help, On phpnuke and phpgroupware.

[HiM]

Jason,
phpNuke problem fixed after downgrade to 4.0.6
Thanks !
And I am still solving to problem with phpgroupware.

Nori
-
- Original Message -
From: "Jason Wong" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, January 10, 2002 10:07 PM
Subject: Re: [PHP-DB] Please help, On phpnuke and phpgroupware.


> On Thursday 10 January 2002 21:53, HiM wrote:
> > Jason,
> > But after downgrade to 4.0.6, is there any new feature I can't use?
> >
> > Nori Chan
>
>
> If you're going to be using only PHPNuke and/or PHPGroupware and they
haven't
> been updated for 4.1.1 then they won't be using any of the new features of
> 4.1.1.
>
> Read the changelog for 4.1.1 to see what's added and fixed.
>
>
> --
> Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
>
> /*
> Let's not complicate our relationship by trying to communicate with each
> other.
> */
>
> --
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Re: [PHP-DB] Please help, On phpnuke and phpgroupware.

[Jason Wong]

On Thursday 10 January 2002 21:53, HiM wrote:
> Jason,
> But after downgrade to 4.0.6, is there any new feature I can't use?
>
> Nori Chan


If you're going to be using only PHPNuke and/or PHPGroupware and they haven't 
been updated for 4.1.1 then they won't be using any of the new features of 
4.1.1.

Read the changelog for 4.1.1 to see what's added and fixed.


-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
Let's not complicate our relationship by trying to communicate with each 
other.
*/

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Re: [PHP-DB] Please help, On phpnuke and phpgroupware.

[HiM]

Jason,
But after downgrade to 4.0.6, is there any new feature I can't use?

Nori Chan

- Original Message -
From: "Jason Wong" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, January 10, 2002 9:41 PM
Subject: Re: [PHP-DB] Please help, On phpnuke and phpgroupware.


> On Thursday 10 January 2002 21:28, HiM wrote:
>
> > I am sorry that last message haven't give the URL...:(
> >
> > http://nori.dns2go.com:8080/phpnuke/html/admin.php
> > http://nori.dns2go.com:8080/phpgroupware/setup/config.php
> >
> > I was trying to use to phpnuke system and also the phpgroupware.
> > I have followed the instruction listed in INSTALL step by step.
> > But I don't know why there's so many error even in phpnuke /
phpgroupware
> >
> > I am using Windows2000 server and apache, php 4.1.1 and mysql 3.23
>
> The recommended php.ini settings for php 4.1.1 are more secure than
previous
> versions BUT will break a lot of the existing PHP applications. Either
change
> the php.ini settings to match those of an earlier version of PHP (eg
4.0.6)
> or wait until the author of the apps in question upgrade them to make it
> compatible with 4.1.1(!)
>
>
> --
> Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
>
> /*
> I don't want people to love me.  It makes for obligations.
> -- Jean Anouilh
> */
>
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Re: [PHP-DB] Please help, On phpnuke and phpgroupware.

[Jason Wong]

On Thursday 10 January 2002 21:28, HiM wrote:

> I am sorry that last message haven't give the URL...:(
>
> http://nori.dns2go.com:8080/phpnuke/html/admin.php
> http://nori.dns2go.com:8080/phpgroupware/setup/config.php
>
> I was trying to use to phpnuke system and also the phpgroupware.
> I have followed the instruction listed in INSTALL step by step.
> But I don't know why there's so many error even in phpnuke / phpgroupware
>
> I am using Windows2000 server and apache, php 4.1.1 and mysql 3.23

The recommended php.ini settings for php 4.1.1 are more secure than previous 
versions BUT will break a lot of the existing PHP applications. Either change 
the php.ini settings to match those of an earlier version of PHP (eg 4.0.6) 
or wait until the author of the apps in question upgrade them to make it 
compatible with 4.1.1(!)


-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
I don't want people to love me.  It makes for obligations.
-- Jean Anouilh
*/

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[PHP-DB] Please help, On phpnuke and phpgroupware.

[HiM]

I am sorry that last message haven't give the URL...:(

http://nori.dns2go.com:8080/phpnuke/html/admin.php
http://nori.dns2go.com:8080/phpgroupware/setup/config.php

I was trying to use to phpnuke system and also the phpgroupware.
I have followed the instruction listed in INSTALL step by step.
But I don't know why there's so many error even in phpnuke / phpgroupware

I am using Windows2000 server and apache, php 4.1.1 and mysql 3.23

Please help , I have to build a groupware tool for my work.

Nori.


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Re: [PHP-DB] Please help asap

[Bogdan Stancescu]

In the printf() line after "5 6 7" you try to print "%s" 24 times but only provide 10 
values. That may have something to do with it...

Bogdan

Jeff Moncrieff wrote:

> Hello
>
>  I am trying  make a script fatch my data form my mysql database. I use
> this script but when execute it is says
> Warning: printf(): too few arguments in
> /home/httpd/html/larken/database.php on line 47
>
> any hints
> Thanks Jeff
>
> 
>
> 
>
> 
> $db = mysql_connect("", "root");
>
> mysql_select_db("larken",$db);
>
> $result = mysql_query("SELECT * FROM custgeninfo",$db);
>
> echo "\n";
>
> echo "Name of contactName of company
> AddressPostal CodeFormat MailEmail
> 
>AddressmodelCommenttelephoneNotesCounty
>
> ID \n";
>
> while ($myrow = mysql_fetch_row($result)) {
>
> //   1 2 34
> 5  6   7
> 
>printf("%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s\n",$myrow[1],$myrow[2],$myrow[3],$myrow[4],$myrow[5],$myrow[6],$myrow[7],$myrow[8],$myrow[9],$myrow[10]);
>
> }
>
> echo ""
>
> ?>
>
> 
>
> 
>
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[PHP-DB] Please help asap

[Jeff Moncrieff]

Hello

 I am trying  make a script fatch my data form my mysql database. I use
this script but when execute it is says
Warning: printf(): too few arguments in
/home/httpd/html/larken/database.php on line 47

any hints
Thanks Jeff









\n";



echo "Name of contactName of company
AddressPostal CodeFormat MailEmail
AddressmodelCommenttelephoneNotesCounty

ID \n";




while ($myrow = mysql_fetch_row($result)) {



//   1 2 34
5  6   7
printf("%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s%s\n",$myrow[1],$myrow[2],$myrow[3],$myrow[4],$myrow[5],$myrow[6],$myrow[7],$myrow[8],$myrow[9],$myrow[10]);








}




echo ""




?>












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Re: [PHP-DB] Please Help Me! thanks!!

[Denny Suvanto]

Try using fsockopen() function :
$file = fsockopen ("www.yahoo.com", 80, $err_num,
$err_msg, 30);

You can find more informations about this fuction in
http://www.php.net/manual/en/function.fsockopen.php

May it help you.

Andy
www.mediahostnet.com

--- G <[EMAIL PROTECTED]> wrote: > Hello! i
would like to useing fopen() fuction to
> open the other Url,
> but i am using a shared server now. When i useing
> fopen() or file()
> fuctions, always have a error message:
> 
> Warning: php_hostconnect: connect failed in
> /usr/xxx/xxx/xxx/html/test.php
> on line 3
> 
> Warning: file("http://www.yahoo.com";) - Bad file
> descriptor in
> /usr/xxx/xxx/xxx/html/test.php on line 3
> 
> How can i do?have any other fuctions can do that?
> Thank You Very Much!!
> 
> 
> 
> 
> 
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>  

__
Do You Yahoo!?
Everything you'll ever need on one web page from News and Sport to Email and Music 
Charts
http://uk.my.yahoo.com

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[PHP-DB] Please help!

[Marios Moutzouris]

 try this :-)

 echo "";
 echo "";
 $row= mysql_fetch_array($result);
 $count  = 0;
 do

  echo "http://www.charmed-guide.com/pictures/ "
.
 $row["pic"] ".jpg\" target=_blank>\n";
  echo "http://www.charmed-guide.com/pictures/"; .
 $row["thumbnail"] . "_th.jpg\"border=0>\n";
  echo "";
  $count++;
  if ($count==2) {
   echo "";
   echo "";
   $count = 0;
  }
 } while ( ( $row = mysql_fetch_array($result) ) );
 echo "";
 echo "";

> ----- Original Message -
> From: "Matt C" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, August 22, 2001 10:09 PM
> Subject: Re: [PHP-DB] Please help!
>
>
> > Can you put it into my code and paste it as a whole? I don't know where
to
> > put it :(
> >
> > _
> > Get your FREE download of MSN Explorer at
http://explorer.msn.com/intl.asp
> >
>


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Re: [PHP-DB] Please help!

[phpnet]

you do not need a counter, just use a 2 column table


| Thats my code!
| 
| What I want to be able to do is have it so instead of one column there are 
| two columns of pictures:
| 
| 
| pic1  | pic2
| pic3  | pic4
| 
| etc..
| 
| 
| How can I do this?
| 
| Thanks in advance!
| 
| 
| 
| 
| http://www.charmed-guide.com/pictures/$row[2].jpg\"; 
| target=_blank>\n");
| printf("http://www.charmed-guide.com/pictures/$row[2]_th.jpg\"; 
| border=0>\n");
| printf("\n");
| 
| $counter++;
| }
| 
| }
| ?>
| 
| _
| Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp
| 
| 
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| 
| 

[EMAIL PROTECTED]

SeaPortNetHosting: Reliable Web Hosting
Plans from $17.95 www.yourname.com
http://www.SeaPortNet.com/
1(209)551-7028 

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Re: [PHP-DB] Please help!

[Justin Buist]

I realize this is trivial... but if you're using things like the code
below alot it can clean things up a bit by using:

if (($counter % 2) == 0 && ($counter != 0))
echo "";
$counter++;
echo "\n";

Just reads easier... less branching, less code.  In an ideal world it'd
optimize into machine code better than the below snippet too -- but that's
get anal retentive about the situtation :)

Justin Buist
Trident Technology, Inc.
4700 60th St. SW, Suite 102
Grand Rapids, MI  49512
Ph. 616.554.2700
Fx. 616.554.3331
Mo. 616.291.2612

On Wed, 22 Aug 2001, Marios Moutzouris wrote:

>
> use that counter variable. check if its equal to 2 before incrementing, if
> it is then "" and reset to zero, otherwise just increment.
>
> if ($counter==2) {
> echo "";
> $counter=0;
> } else {
> $counter++;
> }


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Re: [PHP-DB] Please help!

[Marios Moutzouris]

use that counter variable. check if its equal to 2 before incrementing, if
it is then "" and reset to zero, otherwise just increment.

if ($counter==2) {
echo "";
$counter=0;
} else {
$counter++;
}

marios
- Original Message - Original Message -
From: "Matt C" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 22, 2001 9:37 PM
Subject: [PHP-DB] Please help!


> Thats my code!
>
> What I want to be able to do is have it so instead of one column there are
> two columns of pictures:
>
>
> pic1  | pic2
> pic3  | pic4
>
> etc..
>
>
> How can I do this?
>
> Thanks in advance!
>
>
>
>
> 
> $mysql_access = mysql_connect("", "**", "*");
> mysql_select_db("*", $mysql_access);
>
> $query = "SELECT * FROM Pictures";
> $result = mysql_query($query, $mysql_access);
>
> if(mysql_num_rows($result)) {
>   $counter = 0;
> while($row = mysql_fetch_row($result)) {
>
> printf("http://www.charmed-guide.com/pictures/$row[2].jpg\";
> target=_blank>\n");
> printf("http://www.charmed-guide.com/pictures/$row[2]_th.jpg\";
> border=0>\n");
> printf("\n");
>
> $counter++;
> }
>
> }
> ?>
>
> _
> Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>


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[PHP-DB] Please help!

[Matt C]

Thats my code!

What I want to be able to do is have it so instead of one column there are 
two columns of pictures:


pic1  | pic2
pic3  | pic4

etc..


How can I do this?

Thanks in advance!




http://www.charmed-guide.com/pictures/$row[2].jpg\"; 
target=_blank>\n");
printf("http://www.charmed-guide.com/pictures/$row[2]_th.jpg\"; 
border=0>\n");
printf("\n");

$counter++;
}

}
?>

_
Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp


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RE: [PHP-DB] PLEASE HELP! Problem with ODBC string insert

[Matthew Loff]

It would be a hack, but if the data is only going to be used for the
site (e.g. nobody is going to use the data through access) you could
str_replace() the single quote to something else (perhaps an HTML ASCII
representation like &#;) before the INSERT/UPDATE, and converting
back when SELECTing.  It's ugly, but it would work...


-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 15, 2001 7:21 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert


I tried double quotes and many different combinations of single and
double but nothing has worked.

There HAS to be an answer out there! Do certain SQL statements just not
work in PHP / Access? Is this a bug in Access or PHP?

If I can't get this to work then I won't be able to use PHP for this
site! Someone please shed some light on this!

"Matthew Loff" <[EMAIL PROTECTED]> wrote in message
001701c10dae$53146280$0100a8c0@bang">news:001701c10dae$53146280$0100a8c0@bang...
>
>
> That looks like a valid query to me...  I know this would break SQL92 
> standard, but have you tried enclosing the values in double quotes?  I

> know that's valid with MySQL...
>
> I don't have any other suggestions... Does anyone else?
>
>
> -Original Message-
> From: AKA Hook [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, July 15, 2001 3:56 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
>
>
> Here is the echo output.
>
> UPDATE News SET NewsTitle = 'Text',
> NewsText = 'can\'t use quotes!' WHERE NewsID = 16
>
>
> "Matthew Loff" <[EMAIL PROTECTED]> wrote in message 
> 000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
> >
> > It might help us to see exactly what the query is...
> >
> > Could you insert an echo statement and let us know what SQL query 
> > it's
>
> > producing?
> >
> > echo "
> > INSERT INTO News
> >(NewsDate,NewsTitle,NewsText)
> > VALUES
> > ('$NowDate','$NewsTitle','$NewsText')
> > ";
> >
> >
> > -Original Message-
> > From: AKA Hook [mailto:[EMAIL PROTECTED]]
> > Sent: Sunday, July 15, 2001 1:03 PM
> > To: [EMAIL PROTECTED]
> > Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
> >
> >
> > I am using an Access DBASE and trying to insert text from a form 
> > field. When I use quotes in my text the code errors out, otherwise 
> > it works fine. I have Magic Quotes turned on which is supposed to 
> > fix this problem but IT DOES NOT!
> >
> > I have posted in several forums but no one has been able to help me!

> > Here is my code:
> > 
> > --
> > --
> > 
> > --
> >
> > $NowDate = date("m/d/y");
> >
> > $DBASE = odbc_connect(mno,admin,admin);
> > $getnews = odbc_exec($DBASE,
> > "
> > INSERT INTO News
> >(NewsDate,NewsTitle,NewsText)
> > VALUES
> > ('$NowDate','$NewsTitle','$NewsText')
> > "
> > );
> >
> > Here is the error when using quotes in the "NewsText" form field:
> > 
> > --
> > --
> > 
> > 
> > Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
> > error (missing operator) in query expression ''can\'t use
quotes!')'.,
> > SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
> > line 22
> >
> > Thanks in advance...
> >
> >
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, e-mail: [EMAIL PROTECTED] For 
> > additional commands, e-mail: [EMAIL PROTECTED] To contact 
> > the list administrators, e-mail: [EMAIL PROTECTED]
> >
>
>
>
> --
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> To contact the list administrators, e-mail:
[EMAIL PROTECTED]
>



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Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert

[AKA Hook]

I tried double quotes and many different combinations of single and
double but nothing has worked.

There HAS to be an answer out there! Do certain SQL statements just not work
in PHP / Access? Is this a bug in Access or PHP?

If I can't get this to work then I won't be able to use PHP for this site!
Someone please shed some light on this!

"Matthew Loff" <[EMAIL PROTECTED]> wrote in message
001701c10dae$53146280$0100a8c0@bang">news:001701c10dae$53146280$0100a8c0@bang...
>
>
> That looks like a valid query to me...  I know this would break SQL92
> standard, but have you tried enclosing the values in double quotes?  I
> know that's valid with MySQL...
>
> I don't have any other suggestions... Does anyone else?
>
>
> -Original Message-
> From: AKA Hook [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, July 15, 2001 3:56 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
>
>
> Here is the echo output.
>
> UPDATE News SET NewsTitle = 'Text',
> NewsText = 'can\'t use quotes!' WHERE NewsID = 16
>
>
> "Matthew Loff" <[EMAIL PROTECTED]> wrote in message
> 000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
> >
> > It might help us to see exactly what the query is...
> >
> > Could you insert an echo statement and let us know what SQL query it's
>
> > producing?
> >
> > echo "
> > INSERT INTO News
> >(NewsDate,NewsTitle,NewsText)
> > VALUES
> > ('$NowDate','$NewsTitle','$NewsText')
> > ";
> >
> >
> > -Original Message-
> > From: AKA Hook [mailto:[EMAIL PROTECTED]]
> > Sent: Sunday, July 15, 2001 1:03 PM
> > To: [EMAIL PROTECTED]
> > Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
> >
> >
> > I am using an Access DBASE and trying to insert text from a form
> > field. When I use quotes in my text the code errors out, otherwise it
> > works fine. I have Magic Quotes turned on which is supposed to fix
> > this problem but IT DOES NOT!
> >
> > I have posted in several forums but no one has been able to help me!
> > Here is my code:
> > --
> > --
> > 
> > --
> >
> > $NowDate = date("m/d/y");
> >
> > $DBASE = odbc_connect(mno,admin,admin);
> > $getnews = odbc_exec($DBASE,
> > "
> > INSERT INTO News
> >(NewsDate,NewsTitle,NewsText)
> > VALUES
> > ('$NowDate','$NewsTitle','$NewsText')
> > "
> > );
> >
> > Here is the error when using quotes in the "NewsText" form field:
> > --
> > --
> > 
> > 
> > Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
> > error (missing operator) in query expression ''can\'t use quotes!')'.,
> > SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
> > line 22
> >
> > Thanks in advance...
> >
> >
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > For additional commands, e-mail: [EMAIL PROTECTED]
> > To contact the list administrators, e-mail:
> > [EMAIL PROTECTED]
> >
>
>
>
> --
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> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
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>



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RE: [PHP-DB] PLEASE HELP! Problem with ODBC string insert

[Matthew Loff]

That looks like a valid query to me...  I know this would break SQL92
standard, but have you tried enclosing the values in double quotes?  I
know that's valid with MySQL...  

I don't have any other suggestions... Does anyone else?


-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 15, 2001 3:56 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert


Here is the echo output.

UPDATE News SET NewsTitle = 'Text',
NewsText = 'can\'t use quotes!' WHERE NewsID = 16


"Matthew Loff" <[EMAIL PROTECTED]> wrote in message
000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
>
> It might help us to see exactly what the query is...
>
> Could you insert an echo statement and let us know what SQL query it's

> producing?
>
> echo "
> INSERT INTO News
>(NewsDate,NewsTitle,NewsText)
> VALUES
> ('$NowDate','$NewsTitle','$NewsText')
> ";
>
>
> -----Original Message-
> From: AKA Hook [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, July 15, 2001 1:03 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
>
>
> I am using an Access DBASE and trying to insert text from a form 
> field. When I use quotes in my text the code errors out, otherwise it 
> works fine. I have Magic Quotes turned on which is supposed to fix 
> this problem but IT DOES NOT!
>
> I have posted in several forums but no one has been able to help me! 
> Here is my code:
> --
> --
> 
> --
>
> $NowDate = date("m/d/y");
>
> $DBASE = odbc_connect(mno,admin,admin);
> $getnews = odbc_exec($DBASE,
> "
> INSERT INTO News
>(NewsDate,NewsTitle,NewsText)
> VALUES
> ('$NowDate','$NewsTitle','$NewsText')
> "
> );
>
> Here is the error when using quotes in the "NewsText" form field:
> --
> --
> 
> 
> Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
> error (missing operator) in query expression ''can\'t use quotes!')'.,
> SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
> line 22
>
> Thanks in advance...
>
>
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
> To contact the list administrators, e-mail: 
> [EMAIL PROTECTED]
>



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Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert

[AKA Hook]

Here is the echo output.

UPDATE News SET NewsTitle = 'Text',
NewsText = 'can\'t use quotes!' WHERE NewsID = 16


"Matthew Loff" <[EMAIL PROTECTED]> wrote in message
000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
>
> It might help us to see exactly what the query is...
>
> Could you insert an echo statement and let us know what SQL query it's
> producing?
>
> echo "
> INSERT INTO News
>(NewsDate,NewsTitle,NewsText)
> VALUES
> ('$NowDate','$NewsTitle','$NewsText')
> ";
>
>
> -Original Message-
> From: AKA Hook [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, July 15, 2001 1:03 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
>
>
> I am using an Access DBASE and trying to insert text from a form field.
> When I use quotes in my text the code errors out, otherwise it works
> fine. I have Magic Quotes turned on which is supposed to fix this
> problem but IT DOES NOT!
>
> I have posted in several forums but no one has been able to help me!
> Here is my code:
> 
> 
> --
>
> $NowDate = date("m/d/y");
>
> $DBASE = odbc_connect(mno,admin,admin);
> $getnews = odbc_exec($DBASE,
> "
> INSERT INTO News
>(NewsDate,NewsTitle,NewsText)
> VALUES
> ('$NowDate','$NewsTitle','$NewsText')
> "
> );
>
> Here is the error when using quotes in the "NewsText" form field:
> 
> 
> 
> Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
> error (missing operator) in query expression ''can\'t use quotes!')'.,
> SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
> line 22
>
> Thanks in advance...
>
>
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>



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RE: [PHP-DB] PLEASE HELP! Problem with ODBC string insert

[Matthew Loff]

It might help us to see exactly what the query is...

Could you insert an echo statement and let us know what SQL query it's
producing?

echo "
INSERT INTO News
   (NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
";


-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 15, 2001 1:03 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert


I am using an Access DBASE and trying to insert text from a form field.
When I use quotes in my text the code errors out, otherwise it works
fine. I have Magic Quotes turned on which is supposed to fix this
problem but IT DOES NOT!

I have posted in several forums but no one has been able to help me!
Here is my code:


--

$NowDate = date("m/d/y");

$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
"
INSERT INTO News
   (NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
"
);

Here is the error when using quotes in the "NewsText" form field:



Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
error (missing operator) in query expression ''can\'t use quotes!')'.,
SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
line 22

Thanks in advance...





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[PHP-DB] PLEASE HELP! Problem with ODBC string insert

[AKA Hook]

I am using an Access DBASE and trying to insert text from a form field.
When I use quotes in my text the code errors out, otherwise it works fine.
I have Magic Quotes turned on which is supposed to fix this problem
but IT DOES NOT!

I have posted in several forums but no one has been able to help me!
Here is my code:

--

$NowDate = date("m/d/y");

$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
"
INSERT INTO News
   (NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
"
);

Here is the error when using quotes in the "NewsText" form field:


Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax error
(missing operator) in query expression ''can\'t use quotes!')'., SQL state
37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on line 22

Thanks in advance...





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Re[2]: [PHP-DB] PLEASE HELP !!!

[andRie Is]

I bet, there is something logic error in your program..
PHP+apache is such a small program with low resources that cannot harm your
system ..


mk> ""PHPFAN"" <[EMAIL PROTECTED]> wrote in message
mk> 9eetk0$92d$[EMAIL PROTECTED]">news:9eetk0$92d$[EMAIL PROTECTED]...
>> I am running PHP 4.0.5 on Windows 2000 with SQL server databases.
>> If there is an error with my PHP code, my computer becomes very very slow
>> and  nothing works. The PHP process is using 99 % of the CPU. So I have to
>> restart the computer. This
>> happens everytime there is an error in my PHP code. For the work I am
mk> doing,
>> I have to use PHP with backend SQL
>> Server on Windows Platform.
-- 
Best regards,
 andRiemailto:[EMAIL PROTECTED]



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Re: [PHP-DB] PLEASE HELP !!!

[matthew knight]

when you say an error - do you mean any error?
or specific errors? ie., you're not getting it stuck in an infinite loop?
(cpu time flips up to that when you stick in a loop)

or is it all errors?



""PHPFAN"" <[EMAIL PROTECTED]> wrote in message
9eetk0$92d$[EMAIL PROTECTED]">news:9eetk0$92d$[EMAIL PROTECTED]...
> I am running PHP 4.0.5 on Windows 2000 with SQL server databases.
> If there is an error with my PHP code, my computer becomes very very slow
> and  nothing works. The PHP process is using 99 % of the CPU. So I have to
> restart the computer. This
> happens everytime there is an error in my PHP code. For the work I am
doing,
> I have to use PHP with backend SQL
> Server on Windows Platform.
>
> Any suggestions will be greatly appreciated.
>
> Thank you
> PHPFAN
>
>
>
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[PHP-DB] PLEASE HELP !!!

[PHPFAN]

I am running PHP 4.0.5 on Windows 2000 with SQL server databases.
If there is an error with my PHP code, my computer becomes very very slow
and  nothing works. The PHP process is using 99 % of the CPU. So I have to
restart the computer. This
happens everytime there is an error in my PHP code. For the work I am doing,
I have to use PHP with backend SQL
Server on Windows Platform.

Any suggestions will be greatly appreciated.

Thank you
PHPFAN



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Re: [PHP-DB] Please help me to solve a question

[olinux o]

check the section in the mysql manual on
creating/setting users. i think it's section 6.14. You
probably need to allow users to connect from
connections other than localhost.

olinux


--- David Lee <[EMAIL PROTECTED]> wrote:
> I used MySQL server to create a application database
> and access it by php
> program.
> But now,I can access it only one machine that is
> runing the server.Other
> computer
> can not access it through browser(Internet
> Explorer5.0).My networks is made
> up with
> windows98 "peer to peer" network.Please help me and
> tell me what should i
> do.
> 
> Thank you very much
>   David Lee
> 
> 
> 
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Re: [PHP-DB] Please help me to solve a question

[Yevgeny Dyatlov]

Hello David,
it's not a problem of PHP or MySQL,
it's a problem of your webserver (IIS or Apache or someone else)!
Try to write at adress bar on your broser IP-adress of webserver
mashine or its name in network.
Wednesday, April 25, 2001, 4:45:04 PM, you wrote:

DL> I used MySQL server to create a application database and access it by php
DL> program.
DL> But now,I can access it only one machine that is runing the server.Other
DL> computer
DL> can not access it through browser(Internet Explorer5.0).My networks is made
DL> up with
DL> windows98 "peer to peer" network.Please help me and tell me what should i
DL> do.

DL> Thank you very much
DL>   David Lee






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Best regards,
 Yevgenymailto:[EMAIL PROTECTED]



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[PHP-DB] Please help me to solve a question

[David Lee]

I used MySQL server to create a application database and access it by php
program.
But now,I can access it only one machine that is runing the server.Other
computer
can not access it through browser(Internet Explorer5.0).My networks is made
up with
windows98 "peer to peer" network.Please help me and tell me what should i
do.

Thank you very much
  David Lee



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RE: [PHP-DB] Please help me with this script...

[Mark Roedel]

> -Original Message-
> From: Slider© [mailto:[EMAIL PROTECTED]]
> Sent: Monday, April 02, 2001 9:27 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Please help me with this script...
> 
> 
> Hello,
> 
> I've been working on this script all day, but I'm getting the 
> same failure
> report each time:
> 
> Warning: Supplied argument is not a valid MySQL result resource in
> c:\program files\apache 
> group\apache\htdocs\motoerit\toon_alles.php on line 23
> 
> What could be wrong? This is the code I have:

This usually means there was a problem with your query.  For more info,
including some useable sample code that should help you troubleshoot
exactly what went wrong, see

http://www.php.net/FAQ.php#7.12

---
Mark Roedel ([EMAIL PROTECTED])  ||  "There cannot be a crisis next week.
Systems Programmer / WebMaster  ||   My schedule is already full."
 LeTourneau University  ||-- Henry Kissinger


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[PHP-DB] Please help me with this script...

[Slider©]

Hello,

I've been working on this script all day, but I'm getting the same failure
report each time:

Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\motoerit\toon_alles.php on line
23

What could be wrong? This is the code I have:

line 23 is: while(list($ID, $Naam, $Adres, [..] $Bet_rest, $Overig_info)
= mysql_fetch_array ($result))


---


Boekingen


 
 


");
 print("ID:Naam:Reisdoel:Datum vertrek:Datum terugkomst:Datum betaling voorschot:Datum betaling
rest:  ");

 print("$ID$Naam$Reisdoel$Dat_heen$Dat_
terugWijzigenVerwijderen\n");

 print("");

include ("footer.txt");
?>
---

Thanks in advance!

Kind regards,

Paul Sidler



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RE: [PHP-DB] please help

[Rick Emery]

This error means that mysql is not installed.  Or that the module which
links PHP and mysql is not installed.  I ahd the same error in my RedHat 5.2
machine.  Unfortunately, there was no RPM for to provide this
interconnectivity between PHP and MySQL.  I upgraded to RedHat 7.0;
installed the MySQL, Apache w/PHP, and php-mysql RPMS.  Everything works
like a charm now.

rick
-Original Message-
From: Irwan Agustian [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 12, 2001 11:50 PM
To: Gigi Sze
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] please help




you need password to connect to the mysql
maybe you must check to this URL 
http://hotwired.lycos.com/webmonkey/99/21/index2a_page6.html?tw=programming


regards
 
On Mar 12 Gigi Sze nulis:

> Date: Mon, 12 Mar 2001 14:48:56 +0800
> From: Gigi Sze <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] please help
> 
> hi, everyone ...
> 
> this's my first time to write php program.
> 
> and i've try to execute the following code:
> 
>  
> $link = mysql_connect("localhost", "gigi", "");
> 
> mysql_select_db("cus_db");
> 
> $query = "select count(*) from customer";
> 
> $result = mysql_query($query, $link);
> 
> if ($row=  mysql_fetch_array($result))
> 
> echo "the table cus has ".$row[0]." members.";
> 
> mysq_free_result($result);
> 
> ?>
> 
> but it show:
> 
> Fatal error: Call to unsupported or undefined function mysql_connect() in
/home/httpd/html/test.php3 on line 2
> 
> i don't understand.
> 
> please help
> 
> Gigi
> 


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Re: [PHP-DB] please help

[Irwan Agustian]

you need password to connect to the mysql
maybe you must check to this URL 
http://hotwired.lycos.com/webmonkey/99/21/index2a_page6.html?tw=programming


regards
 
On Mar 12 Gigi Sze nulis:

> Date: Mon, 12 Mar 2001 14:48:56 +0800
> From: Gigi Sze <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] please help
> 
> hi, everyone ...
> 
> this's my first time to write php program.
> 
> and i've try to execute the following code:
> 
>  
> $link = mysql_connect("localhost", "gigi", "");
> 
> mysql_select_db("cus_db");
> 
> $query = "select count(*) from customer";
> 
> $result = mysql_query($query, $link);
> 
> if ($row=  mysql_fetch_array($result))
> 
> echo "the table cus has ".$row[0]." members.";
> 
> mysq_free_result($result);
> 
> ?>
> 
> but it show:
> 
> Fatal error: Call to unsupported or undefined function mysql_connect() in 
>/home/httpd/html/test.php3 on line 2
> 
> i don't understand.
> 
> please help
> 
> Gigi
> 


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Re: [PHP-DB] please help

[Jason Brooke]

> but it show:
>
> Fatal error: Call to unsupported or undefined function mysql_connect() in
/home/httpd/html/test.php3 on line 2
>
> i don't understand.
>
> please help
>
> Gigi

You'll need to read the manual to find out how to enable mysql for your
operating system/version of php
http://www.php.net/manual/en/installation.php

jason




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[PHP-DB] please help

[Gigi Sze]

hi, everyone ...

this's my first time to write php program.

and i've try to execute the following code:



but it show:

Fatal error: Call to unsupported or undefined function mysql_connect() in 
/home/httpd/html/test.php3 on line 2

i don't understand.

please help

Gigi