Error wont go away?

[potter_gold]

 

Hi

 

My first post,  I have an error on a query and do not know how to fix it.  Can anyone help me?

 

Peter

 

It is as follows and appears on this page http://www.freepspwallpaper.co.uk/free-psp-wallpapers/free-babes-psp-wallpapers/

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/freepspw/public_html/dbconnect.php on line 22

 

[fenway]

Post the query that's causing that error-- not the code.

[potter_gold]

So sorry Fenway. 

 

This is the error

 

while ($row = mysql_fetch_array($query)) {
   $catlinks .= ' <img src="bullet.gif" width="7" height="7" border="0" align="center"><a href="viewcategory.php?cid='.$row['wallpaperid'].'">'.$row['categoryid'].'</A><br/>';
}

 

-----

 

This is the rest of the code within my dbconnect file - minus rows for usernames and passwords.

 

// Connect to the database.
$database_connection = mysql_connect(DB_HOST, DB_USERNAME, DB_PASS);
mysql_select_db(DB_NAME);

require('class.pagedresults.php');

// Create the navbar.

// Select what we want.
$sql = 'SELECT * FROM `categories`';

$query = mysql_query($sql);
$catlinks = ' <img src="bullet.gif" width="7" height="7" border="0" align="center"><a HREF="http://www.freepspwallpaper.co.uk">Home</A><br/>';

while ($row = mysql_fetch_array($query)) {
   $catlinks .= ' <img src="bullet.gif" width="7" height="7" border="0" align="center"><a href="viewcategory.php?cid='.$row['wallpaperid'].'">'.$row['categoryid'].'</A><br/>';
}
$template = file_get_contents('template.php');
$template = str_replace('{LINKS}',$catlinks,$template);
?>

[Yesideez]

First - please surround any PHP with PHP tags like this:

echo '[code=php:0] to start and end with 

';[/code]

 

This makes reading source a lot easier.

 

Second, with PHP the error line returned is often not the line in question that has the error.

 

With that type of error message (invalid query) the error won't really be in the mysql_query() call but the line previously somewhere where the query itself is being defined, example:

$query="SELECT * FROM table WHERE field!=field";

 

That's got the error because MySQL uses <> for not equals to and not !=

 

Can you post the query please?