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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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)
);
$que
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'
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 debu
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))
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')";
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
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 quer
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 <[EMA
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
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
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.
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,
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 col
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_
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
rig
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
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 so
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 t
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
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
eptember 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 . "
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 200
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.
> --
t: 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&
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
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 taki
m> cc:
Subject: RE: [PHP-DB] Please help -
Managing relational tables using PHP
he 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 PRO
hanivas 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 &quo
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
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/s
Any ideas how to generate a unique serial number that you can check that
it's valid..
Thanks...
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ooks 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
L 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
>
>
$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 co
-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.
>
>
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
ge-
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
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
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 PROTEC
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
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 A
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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To unsubscribe, vis
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
becom
IL 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 C
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
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 phpnu
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 f
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
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
> Wa
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
Try using fsockopen() function :
$file = fsockopen ("www.yahoo.com", 80, $err_num,
$err_msg, 30);
You can find more informations about this fuction in
http://www.php.net/manual/en/function.fsockopen.php
May it help you.
Andy
www.mediahostnet.com
--- G <[EMAIL PROTECTED]> wrote: > Hello! i
woul
Original Message -
> From: "Matt C" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, August 22, 2001 10:09 PM
> Subject: Re: [PHP-DB] Please help!
>
>
> > Can you put it into my code and paste it as a whole? I don't know where
to
you do not need a counter, just use a 2 column table
| Thats my code!
|
| What I want to be able to do is have it so instead of one column there are
| two columns of pictures:
|
|
| pic1 | pic2
| pic3 | pic4
|
| etc..
|
|
| How can I do this?
|
| Thanks in advance!
|
|
I realize this is trivial... but if you're using things like the code
below alot it can clean things up a bit by using:
if (($counter % 2) == 0 && ($counter != 0))
echo "";
$counter++;
echo "\n";
Just reads easier... less branching, less code. In an ideal world it'd
optimize into machin
om: "Matt C" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 22, 2001 9:37 PM
Subject: [PHP-DB] Please help!
> Thats my code!
>
> What I want to be able to do is have it so instead of one column there are
> two columns of pictures:
>
>
>
Thats my code!
What I want to be able to do is have it so instead of one column there are
two columns of pictures:
pic1 | pic2
pic3 | pic4
etc..
How can I do this?
Thanks in advance!
http://www.charmed-guide.com/pictures/$row[2].jpg\";
target=_blank>\n");
printf("htt
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
gt; 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, J
lto:[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
"Ma
27;$NewsTitle','$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 a
quot;;
-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
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
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 Wi
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$92d$[EMAI
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
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 b
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
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 a
> -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
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:
li
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
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", "");
>
> 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
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
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