Hello, our site is giving this error on user login:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in (path omitted) on line 143

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in (path omitted) on line 150

Our programmer is unavailable - is this something I can fix myself or get help with asap?

Thank you!

You should rename that file to .phps and post the url to it.

There's probably no results from the previous mysql_query select. Possibly something wrong with your database, or how you're selecting it.

I can't really help out without any snippit of code.

what should I snip for you to look at? Thanks.

Here's line 140 - 150:
<?
$sql = "SELECT * FROM mm_newsletter_users WHERE user_id='$mm_id'";
$result = mysql_query ($sql);
while ($row = mysql_fetch_array ($result)) {
$newsletters_active [$row ["newsletter_id"]] = "Y";
}


$sql = "SELECT * FROM mm_newsletters ORDER BY newsletter_id";
$result = mysql_query ($sql);
while ($row = mysql_fetch_array ($result)) {
?>

If tables mm_newsletters and mm_newsletter_users are empty, it's correct behaviour.

To avoid displaying such warning you should change the following:


<?
$sql = "SELECT * FROM mm_newsletter_users WHERE user_id='$mm_id'";
if($result = mysql_query ($sql)) {

while ($row = mysql_fetch_array ($result)) {
$newsletters_active [$row ["newsletter_id"]] = "Y";
}

}

$sql = "SELECT * FROM mm_newsletters ORDER BY newsletter_id";
if ($result = mysql_query ($sql)) {
while ($row = mysql_fetch_array ($result)) {
..
..
..
} // end of while loop

}

?>

It would be also better if you've turned off displaying errors directly on page if it's production environment - it can give sometimes valuable information about php code and/or database structure to potential hackers.

You can do this in two ways:

- Put 'ini_set('display_errors','0');' somewhere at the beginning of every php script

- Set 'display_errors = Off' in php.ini and restart/reload webserver

Originally posted by zoldar
It would be also better if you've turned off displaying errors directly on page if it's production environment - it can give sometimes valuable information about php code and/or database structure to potential hackers.

You can do this in two ways:

- Put 'ini_set('display_errors','0');' somewhere at the beginning of every php script

- Set 'display_errors = Off' in php.ini and restart/reload webserver

or simply put this in the script

error_reporting(0);

and no need to restart the webserver :)

Yes, but I'm not sure if it doesn't turn off logging errors to webserver's logs too... there still should be some way of diagnosing problems

Just add the "@" sign --> @mysql_fetch_array

it should work fine...

if the database is not empty, then you might not be connected to the database..

Peace,

One more thing you can do is to put a statement like this:
echo "SELECT * FROM mm_newsletter_users WHERE user_id='$mm_id'";
before the:
$sql = "SELECT * FROM mm_newsletter_users WHERE user_id='$mm_id'";

To get the querystring that is being used to access your database. Then you can manually go to your database and run the statement to see if it is returning a result.

Placing a @ infront of any statement can suppress the error from being thrown but it might cause undesirable results (for the script). Best to ask your programmer to do a better job at ensuring that all possible errors are handled.

You have MANY problems with your current PHP code.

First off.. do not use short php tags ( i.e <? use <?php instead as this may conflict with XML which also uses <? )

Secondly, you need to create a connection to a Database BEFORE you try to do any mysql_query() calls!

The reason you get invalid resource errors is because it is just that! AN INVALID resource... use mysql_connect() www.php.net and read up on php/mysql