advice on this error

Hi

Can any1 tell me what this error means ?

Thanks.

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/pictures/public_html/functions.php on line 83

PHP:

 

Your query was wrong. Check the syntax, column names etc.

 

also double check your spelling on the result variable.

 

spelling ?

 

Post your code if you have done the above and need further help.

 

Yes, sometimes you can have a typo and not notice it right away... for example:

if your query result is $result but you typed $resut by mistake... you'll get that error.

 

It might not even be the spelling or an incorrect query

- Make sure you aren't passing the actual query string into mysql_fetch_object(), it needs to be a result
- Make sure you are passing the right variable

nico_swd is right though. If you want real clarification post the code that caused the error.

 

ok , here is the code

<?

$sql="select * from members,photos,ratings where members.username=photos.username and photos.photosid=ratings.photosid and numvotes>0 and gender='F'  and approved='Y'  and active=1 order by avgrating desc LIMIT 5";
   $res=mysql_query($sql) OR die(mysql_error());
   $rows = 0;
   ?>

   <?php

   while($obj=mysql_fetch_object($res))
   {


   	if($obj->url=="")
   	{
   		$img="<a href='http://www.***.com/index.php?&username=$obj->username&phid=$obj->photosid''><img border=0 width=50 height=50 src='../pics/$obj->filename'></a>";
   	}
   	else
   	{
   		$img="<a target='_blank' href='$obj->url'><img  border=0 width=100 height=100 src='$obj->url'></a>";
   	}

   	$id=$obj->photosid;
   	$username=$obj->username;
   	print $img;

   	?>
       	<br /><a href="http://www.***.com/index.php?cmd=20&username=<? print $username; ?>">View <? print $username; ?>'s profile</a>
       <?php



   }

   ?>

PHP:

 

what happens if you run the sql statement straight on the server ?(not from the program)

Just to confirm that your query is correct.

 

It works fine, its when I add more sql statements its as if they 'colide' with each other, i.e to much sql.

 

try surrounding table and field names with the ` character

`like`.`this`

sometimes that takes care of the problem... but you should focus on the SQL... that's usually the problem. I doubt that it's a bummed PHP/MYSQL install. You could try it on another server if you think the software might be the problem.