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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.

 

 

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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".  :'(

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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);


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https://forums.phpfreaks.com/topic/68697-solved-error-message/#findComment-345357
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