[SOLVED] error message

[littlepeg]

Hi, everybody.  :)I met a problem and hope you can help me. Any help will be appreciated.

I developed my first php website on my computer and tested it by using my university web domain, and it worked fine. However, when I upload this website to the other web domain that I applied. One of my web page cannot work properly, and shows error message: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/adtsfxfg/public_html/edit_user1.php on line 236. The codes are as follows:

$query = "SELECT user_name, first_name, last_name, address, postcode, tel, email, agreetolist FROM users WHERE user_id=$id";		
$result = @mysql_query ($query); // Run the query.

[color=red]if (mysql_num_rows($result) == 1) { // Valid user ID, show the form.[/color] 

// Get the user's information.
$row = mysql_fetch_array ($result, MYSQL_NUM);

  Would you please tell me why it works on one web domain but can't work on the other web domain. And how can I solve this problems. Thank you.

 

 

[magnetica]

What exactly did you upload?

Just the PHP file?

Did you remember to setup the database and change the connection settings if necessary

[MadTechie]

change

$result = @mysql_query ($query); // Run the query.

 

to

$result = mysql_query ($query) or die(mysql_error()); // Run the query.

 

this should show a more useful error

[littlepeg]

I upload all the files for the website, and it connected to the database and most of the web pages works fine, except this one.  :'(

[MadTechie]

can you try my post above and post the results

[littlepeg]

Hi MadTechie, thank you for your reply. I tried to change the code to what you suggested. The error message is the same:

"Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/adtsfxfg/public_html/edit_user1.php on line 236".  :'(

[MadTechie]

Humm

 

you didn't get an extra message ie

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in

 

also you did use exaclty what i posted (removing the @ as well)

can you post what the following outputs

 

$query = "SELECT user_name, first_name, last_name, address, postcode, tel, email, agreetolist FROM users WHERE user_id=$id";
echo $query;
$result = mysql_query ($query) or die(mysql_error()); // Run the query.
$x = mysql_num_rows($result);

if ($x == 1) { // Valid user ID, show the form.

// Get the user's information.
$row = mysql_fetch_array ($result, MYSQL_NUM);

[MadTechie]

as a side note, i assume you have the

<?php

$link = mysql_connect("localhost", "mysql_user", "mysql_password");

mysql_select_db("db", $link);

?>

above the code posted

[littlepeg]

Hi MadTechie, I just noticed I put the code you suggested in the wrong place. I changed it in the right place now and the error is: Unknown column 'email' in 'field list'.

[MadTechie]

check the database for the field "email",

then either add it or change it to email OR change the query to the one from the database

[littlepeg]

Hi MadTechie, it works now. Silly mistake. Thank you very much.  :)

[MadTechie]

your welcome

as a note prefixing the @ omits the error (so use wisely) also the

or die(mysql_error())

is VERY useful, keep them inmind,

 

can you click solved (bottom left) if all is well