RE: [PHP-DB] Update data problem
Ok, I was trying to keep it simple :), but here goes:
In essence the problem is that (at the end of the script)
income_out is coming out as just $option_close_price and not the
intended $option_close_price*$no_sold, and
profit_loss is coming out as just -$option_close_price and not the
intended $income_in-$option_close_price*$no_sold.
Justin
\n";
print "IDOpen
DateShareCodeShort or Long
TradeExpiryExcerciseOption
PriceNumber PurchasedNumber SoldIncome
InIncome Out";
while ($row = mysql_fetch_array($result))
{
print "";
print "";
print "";
print $row["id"];
print "";
print $row["open_date"];
print "";
print $row["share"];
print "";
print $row["code"];
print "";
print $row["short_long_trade"];
print "";
print $row["expiry"];
print "";
print $row["excercise"];
print "";
print $row["option_price"];
print "";
print $row["no_purchased"];
print "";
print $row["no_sold"];
print "";
print $row["income_in"];
print "";
print $row["income_out"];
print "\n";
}
print "\n";
?>
Please enter closing details;
Closed Date:
Option Close Price
Notes
-Original Message-
From: Sam Chill [mailto:[EMAIL PROTECTED]
Sent: Thursday, 24 June 2004 3:25 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Update data problem
Justin,
Is the code you posted all of the code or just a small snippet? You
don't appear to be setting $C or $G. Could you post a larger code dump
please?
-Sam
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Re: [PHP-DB] Update data problem
Justin, Is the code you posted all of the code or just a small snippet? You don't appear to be setting $C or $G. Could you post a larger code dump please? -Sam -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Update data problem
Sam, I have changed it but the result is still the same: Basically I am getting $E = $C , it seems that it is not picking up the value of $G (and there is a value of 500 in there and it is the only row of data). -Original Message- From: Sam Chill [mailto:[EMAIL PROTECTED] Sent: Thursday, 24 June 2004 2:55 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Update data problem $E= "$C*$G"; Should be: $E = $C * $G; Never use quotation marks when doing math. Quotation marks are used for strings. Hope that helps. -- 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] Update data problem
$E= "$C*$G"; Should be: $E = $C * $G; Never use quotation marks when doing math. Quotation marks are used for strings. Hope that helps. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Update data problem
Hi there, I hoped someone might be able to tell me why this is not
working.
I have a form that updates the table 'media' with the 3 values below,
that and some other variables are the only data in the table. That works
nicely.
I then want to update one of the other columns ($E) based on the data
I've just submitted - that is the multiplication of the two variables. I
then want to input this variable back into the table. What happens is
that the new values are used ($C), but the original value in the table
($G) is ignored. Why is this? I would have thought it was pretty
straight-froward to extract 2 values from a table, multiply them
together and then input it back as a new variable.
Thanks
Justin
if($update)
{
mysql_pconnect("localhost","root","password");
mysql_select_db("options");
$query = "UPDATE media SET A='$A', B='$B', C='$C'";
$result = mysql_query($query);
$query = "SELECT C, G FROM media";
$result = mysql_query($query);
$E= "$C*$G";
$query = "UPDATE media SET E='$E'";
$result = mysql_query($query);
}
Re: [PHP-DB] value error in PHP form
The reason for this is due to the variable "$action" not having any relation
to the action. Therefore, you should use the following if/then statement:
if ($submit !== '') {
// STATEMENTS
}
The "submit" name/value pair variable passed via your form is a named
variable that possesses the value of "Submit Query", so you could
essentially make the if/then statement be:
if ($submit == 'Submit Query') {
// STATEMENTS
}
However, you will have to ensure that the "register_globals" directive has a
value of "Yes" in your PHP.ini configuration file.
If you're unsure of the value of your "register_globals" directive, create
and execute the following script:
This function will state all of the environment variables associated with
your install of PHP on your server.
Hope this helps...
-Zak
On 6/20/04 5:18 AM, "Rinku" <[EMAIL PROTECTED]> wrote:
> Actually I want to use the function like
> if($action="Login")
> {
> Statements;
> }
> But here I am not getting any value in $action even I click on it.
> When I was using Linux at that time I had not this kind of problem.
>
> Rinku
> Marvin Hechanova <[EMAIL PROTECTED]> wrote:
> You have to assign values to your name and action
> e.g.
>
> [input] " value=Rinku>
> [input] " type=submit value="Submit Query"
> value="Login">
>
> but why would you want to print the results?
>
>
> On Sun, 20 Jun 2004 04:22:59 -0700 (PDT), Rinku wrote:
>>
>> Dear All,
>>
>> I have installed PHP on WinXp. I am using MySql as Backend on Apache server.
>> Now the problem is..
>> I am writing this code :
>>
>>
>>
>>
>>
>>
>>
>>
>> [input]
>> [input]
>>> print $action;
>> print $Name;
>> ?>
>>
>>
>>
>> Here I should get output as LoginRinku
>> But I am getting nothing.
>>
>> Can any of you guide me on this?
>>
>> Regards,
>> Rinku
>>
>>
>> -
>> Do you Yahoo!?
>> New and Improved Yahoo! Mail - Send 10MB messages!
>
>
> -
> Do you Yahoo!?
> Yahoo! Mail - 50x more storage than other providers!
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Re: [PHP-DB] value error in PHP form
Hi Rinku, One problem I see is that the HTML tags appear to be off... Your HTML is improperly formatted, which could be causing rendering anomalies to occur, in addition to the register globals issue which I'll mention shortly. Please correct your HTML code to read: YOUR_PAGE_TITLE_HERE HTML_BODY_HERE Two, Like Kenny said, change the "register_globals = Off" directive in your PHP.ini file to " register_globals = On" so that variables are carried from your form values into your post-processing PHP print functions. Hope this helps, Best, -Zak On 6/20/04 4:34 AM, "Kenny" <[EMAIL PROTECTED]> wrote: > This could be due to register globals on/off > > Use > print $_POST['action']; > print $_POST['Name']; > > Kenny > > -Original Message- > From: Rinku [mailto:[EMAIL PROTECTED] > Sent: 20 June 2004 12:23 > To: [EMAIL PROTECTED] > Subject: [PHP-DB] value error in PHP form > > Dear All, > > I have installed PHP on WinXp. I am using MySql as Backend on Apache > server. > Now the problem is.. > I am writing this code : > > > > > > > > > > > > > > print $action; > print $Name; > ?> > > > > Here I should get output as LoginRinku > But I am getting nothing. > > Can any of you guide me on this? > > Regards, > Rinku > > > > > > > - > Do you Yahoo!? > New and Improved Yahoo! Mail - Send 10MB messages! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: problem....
I may be incorrect, but I do not think that by "redefining" the
variable $pic is destroying its old contents. If you are familiar with
how programming works, you realize that the "stuff" on the right side
of the "=" sign is evaluated first, and then assigned to the left side
of the "=" sign. Because $pic is in the mysql_fetch_array($pic) (on the
right), $pic is evaluated before it ever gets "destroyed" by the
redefining.
PHP may not be this way, but I think that it is. If I am incorrect,
please let me know. I hope this helps out (with something).
~Philip
On Jun 23, 2004, at 11:47 AM, Cole S. Ashcraft wrote:
Kim,
Won't redefining the variable destroy the MySQL Resource, resulting in
a resource invalid error, like he has?
$pic = mysql_query($sql,$connection); //Defines variable
$pic //Redefining variable, dumps old contents = //calling on an
invalid MySQL Resource, because it doesn't exist
mysql_fetch_array($pic);
My comments are in green.
Cole
Kim Steinhaug wrote:
Well, your query looked confusing to me, but break it up like this
First :
select
Then, name your coloumns, seperate with commas, or use wildcard for
all
*
Define from what table you need to select
FROM database_table
Add some select statements with WHERE
WHERE
and the statements itself
1=1
What did I just write now? Looking at your example I guess :
1. you need the coloumns : Rune, username
2. your table : RuneRunner
3. your select criteria : UserID=3
I would then write your SQL query as this :
$sql = "SELECT Rune, username FROM RuneRunner WHERE UserID=3";
Then you could do your code,
$pic = mysql_query($sql,$connection);
$pic = mysql_fetch_array($pic);
You would also want to add a database abstraction layer, as for
debugging
purposes this will make your life much easier.
Your original code looks to correct if you rewrite it abit, like this
:
SELECT RuneRunner_1.Rune, RuneRunner_1.username FROM RuneRunner
RuneRunner_1
WHERE (RuneRunner_1.User_ID=3);
Not tested but it should work fine. On the other hand, the
referencename
after the
table name should be a short one to make the statement much easier to
read,
somthing like :
SELECT rr.Rune, rr.username FROM RuneRunner rr WHERE (rr.User_ID=3);
Then again, since you only have one table in your statement there is
no need
for the reference name,
and since your WHERE clause is a single statement there is no need
for the
paranthese either.
So you are still back to :
SELECT Rune, username FROM RuneRunner WHERE User_ID=3;
--
--
Kim Steinhaug
--
There are 10 types of people when it comes to binary numbers:
those who understand them, and those who don't.
--
www.steinhaug.com - www.easywebshop.no - www.webkitpro.com
--
"Water_foul" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
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Re: [PHP-DB] SQL help
I have tried that and I don't get an error, but I don't get any records returned either. And I have lowered the search string like you mentioned. Here's what I tried ( Access syntax ): SELECT autoQuesID,fldQuesTitle,fldBody FROM tblFAQ_Question WHERE LCase(fldBody) LIKE '%$strSearchFor%'; Nicole Swan wrote: Have you tried lowering the fldBody as well? Like: SELECT autoQuesID,fldQuesTitle,fldBody FROM tblFAQ_Question WHERE LOWER(fldBody) LIKE '%$strSearchFor%'; And $strSearchFor has already been lowered, of course. --Nicole --- Nicole Swan Web Programming Specialist Carroll College CCIT (406)447-4310 -Original Message- From: Gabe [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 23, 2004 8:59 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] SQL help I'm using PHP with ADOdb ( and an MS Access 2000 db ) to write a simple SQL statement but was running into some case sensitivity issues. Here's my SQL currently: SELECT autoQuesID,fldQuesTitle,fldBody FROM tblFAQ_Question WHERE fldBody LIKE '%$strSearchFor%'; All I'm trying to do is have the users search string searched for in the "fldBody" field. However, I'm having problems trying to get it so that the search is case-insensitive. For instance: If I search on "Airline", I get 1 record. If I search on "airline", I get 0 records. I make the value of $strSearchFor lower case ( using strtolower() ), but I don't know how to get it so that the contents of the "fldBody" field is lower case also. I can't seem to find any functions or operators that remove the case-sensitivity. Any help would be much appreciated! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: problem....
Kim,
Won't redefining the variable destroy the MySQL Resource, resulting in a
resource invalid error, like he has?
$pic = mysql_query($sql,$connection); //Defines variable
$pic //Redefining variable, dumps old contents = //calling on an invalid MySQL Resource, because it doesn't exist mysql_fetch_array($pic);
My comments are in green.
Cole
Kim Steinhaug wrote:
Well, your query looked confusing to me, but break it up like this
First :
select
Then, name your coloumns, seperate with commas, or use wildcard for all
*
Define from what table you need to select
FROM database_table
Add some select statements with WHERE
WHERE
and the statements itself
1=1
What did I just write now? Looking at your example I guess :
1. you need the coloumns : Rune, username
2. your table : RuneRunner
3. your select criteria : UserID=3
I would then write your SQL query as this :
$sql = "SELECT Rune, username FROM RuneRunner WHERE UserID=3";
Then you could do your code,
$pic = mysql_query($sql,$connection);
$pic = mysql_fetch_array($pic);
You would also want to add a database abstraction layer, as for debugging
purposes this will make your life much easier.
Your original code looks to correct if you rewrite it abit, like this :
SELECT RuneRunner_1.Rune, RuneRunner_1.username FROM RuneRunner RuneRunner_1
WHERE (RuneRunner_1.User_ID=3);
Not tested but it should work fine. On the other hand, the referencename
after the
table name should be a short one to make the statement much easier to read,
somthing like :
SELECT rr.Rune, rr.username FROM RuneRunner rr WHERE (rr.User_ID=3);
Then again, since you only have one table in your statement there is no need
for the reference name,
and since your WHERE clause is a single statement there is no need for the
paranthese either.
So you are still back to :
SELECT Rune, username FROM RuneRunner WHERE User_ID=3;
--
--
Kim Steinhaug
--
There are 10 types of people when it comes to binary numbers:
those who understand them, and those who don't.
--
www.steinhaug.com - www.easywebshop.no - www.webkitpro.com
--
"Water_foul" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
--
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[PHP-DB] Re: problem....
Well, your query looked confusing to me, but break it up like this
First :
select
Then, name your coloumns, seperate with commas, or use wildcard for all
*
Define from what table you need to select
FROM database_table
Add some select statements with WHERE
WHERE
and the statements itself
1=1
What did I just write now? Looking at your example I guess :
1. you need the coloumns : Rune, username
2. your table : RuneRunner
3. your select criteria : UserID=3
I would then write your SQL query as this :
$sql = "SELECT Rune, username FROM RuneRunner WHERE UserID=3";
Then you could do your code,
$pic = mysql_query($sql,$connection);
$pic = mysql_fetch_array($pic);
You would also want to add a database abstraction layer, as for debugging
purposes this will make your life much easier.
Your original code looks to correct if you rewrite it abit, like this :
SELECT RuneRunner_1.Rune, RuneRunner_1.username FROM RuneRunner RuneRunner_1
WHERE (RuneRunner_1.User_ID=3);
Not tested but it should work fine. On the other hand, the referencename
after the
table name should be a short one to make the statement much easier to read,
somthing like :
SELECT rr.Rune, rr.username FROM RuneRunner rr WHERE (rr.User_ID=3);
Then again, since you only have one table in your statement there is no need
for the reference name,
and since your WHERE clause is a single statement there is no need for the
paranthese either.
So you are still back to :
SELECT Rune, username FROM RuneRunner WHERE User_ID=3;
--
--
Kim Steinhaug
--
There are 10 types of people when it comes to binary numbers:
those who understand them, and those who don't.
--
www.steinhaug.com - www.easywebshop.no - www.webkitpro.com
--
"Water_foul" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> why doesn't this work:
> $pic=mysql_query('SELECT Rune, username FROM RuneRunner
> RuneRunner_1 WHERE (User_ID = 3)',$connection);
> $pic=mysql_fetch_array($pic);
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RE: [PHP-DB] SQL help
Have you tried lowering the fldBody as well? Like: SELECT autoQuesID,fldQuesTitle,fldBody FROM tblFAQ_Question WHERE LOWER(fldBody) LIKE '%$strSearchFor%'; And $strSearchFor has already been lowered, of course. --Nicole --- Nicole Swan Web Programming Specialist Carroll College CCIT (406)447-4310 -Original Message- From: Gabe [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 23, 2004 8:59 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] SQL help I'm using PHP with ADOdb ( and an MS Access 2000 db ) to write a simple SQL statement but was running into some case sensitivity issues. Here's my SQL currently: SELECT autoQuesID,fldQuesTitle,fldBody FROM tblFAQ_Question WHERE fldBody LIKE '%$strSearchFor%'; All I'm trying to do is have the users search string searched for in the "fldBody" field. However, I'm having problems trying to get it so that the search is case-insensitive. For instance: If I search on "Airline", I get 1 record. If I search on "airline", I get 0 records. I make the value of $strSearchFor lower case ( using strtolower() ), but I don't know how to get it so that the contents of the "fldBody" field is lower case also. I can't seem to find any functions or operators that remove the case-sensitivity. Any help would be much appreciated! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] SQL help
I'm using PHP with ADOdb ( and an MS Access 2000 db ) to write a simple SQL statement but was running into some case sensitivity issues. Here's my SQL currently: SELECT autoQuesID,fldQuesTitle,fldBody FROM tblFAQ_Question WHERE fldBody LIKE '%$strSearchFor%'; All I'm trying to do is have the users search string searched for in the "fldBody" field. However, I'm having problems trying to get it so that the search is case-insensitive. For instance: If I search on "Airline", I get 1 record. If I search on "airline", I get 0 records. I make the value of $strSearchFor lower case ( using strtolower() ), but I don't know how to get it so that the contents of the "fldBody" field is lower case also. I can't seem to find any functions or operators that remove the case-sensitivity. Any help would be much appreciated! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] problem....
Please post the table definition to the mailing list so we can help you
more quickly. Also, is there a table called RuneRunner_1 or are you
meaning to use RuneRunner_1 as an alias?
Odds are, your problem is not PHP related, it's an error in your SQL syntax.
Jon
water_foul wrote:
i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is
there a icq chatroom for php?)
"Shahmat Bin Dahlan" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Your SQL statement:
'SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)'
What is "RuneRunner" and "RuneRunner_1"? Are these two different tables.
If it is, you might want to separate them with a comma.
Why not try getting rid of the brackets surrounding "User_ID=3"? And
also wrap single quotes around the digit "3".
- Original Message -
From: "water_foul" <[EMAIL PROTECTED]>
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: [PHP-DB] problem
I checked em all they were right
"Daniel Clark" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Sounds like it doesn't like your SQL statement. Perhaps a field
or table
name is incorrect?
Warning: mysql_fetch_array(): supplied argument is not a valid
MySQL> > result
resource in ...
"Daniel Clark" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]>
What error are you getting?
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM
RuneRunner> >> > RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
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Re[2]: [PHP-DB] problem....
Hello water_foul,
what about this?
If you do for the first time: $pic=mysql_fetch_array($pic) it will
work, but the SECOND TIME that you do:
$pic=mysql_fetch_array($pic) it will fail, and thats because $pic
is no longer a resource identificator... and you will get:
> > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> > result
>
try this:
$pic=mysql_query("Select Rune, username FROM runeRunner where
(User_ID=3)0,$connection);
$xx=mysql_fetch_array($pic);
do you follow me?...
> >> > why doesn't this work:
> >> > $pic=mysql_query('SELECT Rune, username FROM RuneRunner
> >> > RuneRunner_1 WHERE (User_ID = 3)',$connection);
> >> > $pic=mysql_fetch_array($pic);
P.S. if you need to trace, you can do var_dump($pic);
--
Best regards,
Pablo
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RE: [PHP-DB] problem....
Haven't seen that this has been solved yet so I thought I'd throw my two
cents into the fray. First of all, I'd second the earlier suggestion that
you assign your query to a variable. And, second, I'd recommend that you
echo out your query to the browser just before it is submitted to the
database server. That way, you can actually _see_ what's being sent to the
server. I have found this to be an invaluable troubleshooting tactic and it
may help you find the root cause of your invalid MySQL result.
> -Original Message-
> From: David Robley [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, June 23, 2004 2:53 AM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] problem
>
>
> Water_foul wrote:
>
> > i fixed those things and it didn't fix it :( :( :( :( :( :(
> :( :( :( is
> > there a icq chatroom for php?)
> > "Shahmat Bin Dahlan" <[EMAIL PROTECTED]> wrote in message
> > news:[EMAIL PROTECTED]
> >> Your SQL statement:
> >>
> >> 'SELECT Rune, username FROM RuneRunner
> >> RuneRunner_1 WHERE (User_ID = 3)'
> >>
> >> What is "RuneRunner" and "RuneRunner_1"? Are these two
> different tables.
> >> If it is, you might want to separate them with a comma.
> >>
> >> Why not try getting rid of the brackets surrounding
> "User_ID=3"? And
> >> also wrap single quotes around the digit "3".
> >>
> >>
> >> - Original Message -
> >> From: "water_foul" <[EMAIL PROTECTED]>
> >> Date: Wednesday, June 23, 2004 8:58 am
> >> Subject: Re: [PHP-DB] problem
> >>
> >> > I checked em all they were right
> >> > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message
> >> >
> news:[EMAIL PROTECTED]
> >> > > Sounds like it doesn't like your SQL statement.
> Perhaps a field
> >> > or table
> >> > > name is incorrect?
> >> > >
> >> > > > Warning: mysql_fetch_array(): supplied argument is
> not a valid
> >> > MySQL> > result
> >> > > > resource in ...
> >> > > > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message
> >> > > >
> >> >
> news:[EMAIL PROTECTED]>
> >> >> What error are you getting?
> >> > > >>
> >> > > >> > why doesn't this work:
> >> > > >> > $pic=mysql_query('SELECT Rune, username FROM
> >> > RuneRunner> >> > RuneRunner_1 WHERE (User_ID =
> 3)',$connection);
> >> > > >> > $pic=mysql_fetch_array($pic);
> >> >
>
> Two suggestions for debugging mysql problems:
>
> 1) Create your query as a variable so you can echo it and see
> exactly what
> is being passed to mysql
>
> $query = "SELECT Rune, username FROM RuneRunner, RuneRunner_1
> WHERE User_ID
> = 3";
>
> 2) Use mysql_error() to get the actual error from mysql
> $result = mysql_query($query,$connection);
> echo mysql_error();
> $pic=mysql_fetch_array($result);
>
> --
> David Robley
>
> "I'd like to learn a new card game," Tom said wistfully.
>
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>
>
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Re: [PHP-DB] Value Error in execution
>Dear all,
>Can you aswer for the Query as follow :
>if(isset(_POST['VARIABLE1']$$(_POST['VARIABLE2'])
>{>print "testing";
>STATEMENTS;
>}
I'm not sure how this is not throwing a fatal error. Why do you have two
$'s in the middle and no $ preceeding variable names?
does this work?
if(isset($_POST['VARIABLE1']) && isset($_POST['VARIABLE2']) ) {
print "testing";
}
HTH
Jeff
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[PHP-DB] Value Error in execution
Dear all,
Can you aswer for the Query as follow :
if(isset(_POST['VARIABLE1']$$(_POST['VARIABLE2'])
{
print "testing";
STATEMENTS;
}
Here, even I havent set the value for the variables then even I am getting the output
like "testing".
Can any of you suggest me the sollution for the query ?
Thanks,
Rinku
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[PHP-DB] Re: Multiple Inserts
Solved... I needed to find a way to get the latest auto_increment value.
I did this using,
$new_id = mysql_insert_id();
Thanks,
Declan.
Thanks for the suggestions. I have
$TimeSheetID=$_POST['TimeSheetID'];
at the start of the script. However, what I neglected to mention is that
TimeSheetID is auto_increment, in first table, tblTimesheet.
I will try the other suggestions, and respond with my results.
Thanks,
Declan.
---
Good point. Since it's form data, what about $_POST['TimesheetID'] ?
> Don't see anything obviously wrong with your query string. Are the inserts
> happening in the same block of code, i.e., are you sure that
> _$TimesheetID_ has a value in it when you're performing the second insert?
>
>
> -dave
>
>
>> I am using a form to Insert data into 2 tables in the same database.
>
>> $TimesheetID needs to be in each table. However, it is not being
> inserted
>> into the second table, "tblTimesheetDetails" . Any advise?
>
>> $result_timesheet=mysql_query("INSERT INTO tblTimesheet (TimesheetID,
>> WorkerID, ClientID, TimesheetDate, ProspectiveOrRetrospective) VALUES
>>
>
('$TimeSheetID','$WorkerID','$ClientID','$TimesheetDate','$ProspectiveOrRetr
>> ospective')")or die("Insert Error: ".mysql_error());
>
>> $result_timesheetdetails=mysql_query("INSERT INTO tblTimesheetDetails
>> (TimesheetID, ActivityTypeID, TimeSpentHours, TimeSpentMinutes) VALUES
>>
>
('$TimeSheetID','$ActivityTypeID','$TimeSpentHours','$TimeSpentMinutes')")or
>> die("Insert Error: ".mysql_error());
>
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Re: [PHP-DB] adding a mysql query as text into a history table... ?
thanx guys. "Jason Wong" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > On Wednesday 23 June 2004 17:14, Gawie Marais wrote: > > > i would like to add the whole query into a text field in the 'history' > > table, but i get errors when i try to add it... can anyone help ? > > > > this is the error :- > > > > Invalid query: You have an error in your SQL syntax near '2004-06-23 > > 11:09:57', date_modified = '-00-00 00:00:00', date_closed =' at line 6 > > You'll need to escape the string containing the query before you can insert > it. > > -- > Jason Wong -> Gremlins Associates -> www.gremlins.biz > Open Source Software Systems Integrators > * Web Design & Hosting * Internet & Intranet Applications Development * > -- > Search the list archives before you post > http://marc.theaimsgroup.com/?l=php-db > -- > /* > If you always postpone pleasure you will never have it. Quit work and play > for once! > */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Multiple Inserts
Thanks for the suggestions. I have
$TimeSheetID=$_POST['TimeSheetID'];
at the start of the script. However, what I neglected to mention is that TimeSheetID
is auto_increment, in first table, tblTimesheet.
I will try the other suggestions, and respond with my results.
Thanks,
Declan.
---
Good point. Since it's form data, what about $_POST['TimesheetID'] ?
> Don't see anything obviously wrong with your query string. Are the inserts
> happening in the same block of code, i.e., are you sure that
> _$TimesheetID_ has a value in it when you're performing the second insert?
>
>
> -dave
>
>
>> I am using a form to Insert data into 2 tables in the same database.
>
>> $TimesheetID needs to be in each table. However, it is not being
> inserted
>> into the second table, "tblTimesheetDetails" . Any advise?
>
>> $result_timesheet=mysql_query("INSERT INTO tblTimesheet (TimesheetID,
>> WorkerID, ClientID, TimesheetDate, ProspectiveOrRetrospective) VALUES
>>
> ('$TimeSheetID','$WorkerID','$ClientID','$TimesheetDate','$ProspectiveOrRetr
>> ospective')")or die("Insert Error: ".mysql_error());
>
>> $result_timesheetdetails=mysql_query("INSERT INTO tblTimesheetDetails
>> (TimesheetID, ActivityTypeID, TimeSpentHours, TimeSpentMinutes) VALUES
>>
> ('$TimeSheetID','$ActivityTypeID','$TimeSpentHours','$TimeSpentMinutes')")or
>> die("Insert Error: ".mysql_error());
>
Re: [PHP-DB] adding a mysql query as text into a history table... ?
On Wednesday 23 June 2004 17:14, Gawie Marais wrote: > i would like to add the whole query into a text field in the 'history' > table, but i get errors when i try to add it... can anyone help ? > > this is the error :- > > Invalid query: You have an error in your SQL syntax near '2004-06-23 > 11:09:57', date_modified = '-00-00 00:00:00', date_closed =' at line 6 You'll need to escape the string containing the query before you can insert it. -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* If you always postpone pleasure you will never have it. Quit work and play for once! */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] adding a mysql query as text into a history table... ?
hi guys, i have created a 'history' table and would like to insert the 'bare' sql query into that table just to check who did what on the database... now, i have this :- $CustUpdate = "INSERT INTO record SET recid = '', customer_code = '$varCustomerCode', id_reg = '$varId_Reg', date_added = '$getCurrentDate', date_modified = '-00-00 00:00:00', date_closed = '-00-00 00:00:00', technician = '$varConsultant', flag = '$flag', contact_person = '$varContactPerson', contact_number = '$varContactNumber', contact_email = '$varContactEmail', problem_type = '$varProblemType', problem_description = '$varProblemDescription', priority = '$varPriority', technical_feedback = '', customer_name = '$varCustomerName' "; i would like to add the whole query into a text field in the 'history' table, but i get errors when i try to add it... can anyone help ? this is the error :- Invalid query: You have an error in your SQL syntax near '2004-06-23 11:09:57', date_modified = '-00-00 00:00:00', date_closed =' at line 6 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Multiple Inserts
some tips when inserting data:
instead of inserting into the the table, do some debugging,echo the insert
string and check if:
1- all string fields are between single comma ('field_name').Numéric fields
do not need single comma;
2- all not null fields are present in the field list;
3- you're respecting the fields order (insert into t (x,y,z) values
(val_x,val_y,val_z) );
hope it helps.
Rui Cunha
Daniel Clark writes:
Any errors? Is the all the other data inserting into the second table?
Hello,
I am using a form to Insert data into 2 tables in the same database.
$TimesheetID needs to be in each table. However, it is not being inserted
into the second table, "tblTimesheetDetails" . Any advise?
$result_timesheet=mysql_query("INSERT INTO tblTimesheet (TimesheetID,
WorkerID, ClientID, TimesheetDate, ProspectiveOrRetrospective) VALUES
('$TimeSheetID','$WorkerID','$ClientID','$TimesheetDate','$ProspectiveOrRetr
ospective')")or die("Insert Error: ".mysql_error());
$result_timesheetdetails=mysql_query("INSERT INTO tblTimesheetDetails
(TimesheetID, ActivityTypeID, TimeSpentHours, TimeSpentMinutes) VALUES
('$TimeSheetID','$ActivityTypeID','$TimeSpentHours','$TimeSpentMinutes')")or
die("Insert Error: ".mysql_error());
Thanks,
Declan.
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Re: [PHP-DB] problem....
Water_foul wrote:
> i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is
> there a icq chatroom for php?)
> "Shahmat Bin Dahlan" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
>> Your SQL statement:
>>
>> 'SELECT Rune, username FROM RuneRunner
>> RuneRunner_1 WHERE (User_ID = 3)'
>>
>> What is "RuneRunner" and "RuneRunner_1"? Are these two different tables.
>> If it is, you might want to separate them with a comma.
>>
>> Why not try getting rid of the brackets surrounding "User_ID=3"? And
>> also wrap single quotes around the digit "3".
>>
>>
>> - Original Message -
>> From: "water_foul" <[EMAIL PROTECTED]>
>> Date: Wednesday, June 23, 2004 8:58 am
>> Subject: Re: [PHP-DB] problem
>>
>> > I checked em all they were right
>> > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message
>> > news:[EMAIL PROTECTED]
>> > > Sounds like it doesn't like your SQL statement. Perhaps a field
>> > or table
>> > > name is incorrect?
>> > >
>> > > > Warning: mysql_fetch_array(): supplied argument is not a valid
>> > MySQL> > result
>> > > > resource in ...
>> > > > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message
>> > > >
>> > news:[EMAIL PROTECTED]>
>> >> What error are you getting?
>> > > >>
>> > > >> > why doesn't this work:
>> > > >> > $pic=mysql_query('SELECT Rune, username FROM
>> > RuneRunner> >> > RuneRunner_1 WHERE (User_ID = 3)',$connection);
>> > > >> > $pic=mysql_fetch_array($pic);
>> >
Two suggestions for debugging mysql problems:
1) Create your query as a variable so you can echo it and see exactly what
is being passed to mysql
$query = "SELECT Rune, username FROM RuneRunner, RuneRunner_1 WHERE User_ID
= 3";
2) Use mysql_error() to get the actual error from mysql
$result = mysql_query($query,$connection);
echo mysql_error();
$pic=mysql_fetch_array($result);
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David Robley
"I'd like to learn a new card game," Tom said wistfully.
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