[xxvatarxx]

i have a php file and its throwing up these errors

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/xxvatarx/public_html/games/cats.php on line 5


Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/xxvatarx/public_html/games/cats.php on line 13




line 5 =

PHP Code:

$cat = mysql_fetch_object($qr1); 

line 13 =

PHP Code:

while( $row = mysql_fetch_object($qr1) ){ 

Any ideas?


this is the code of the page

PHP Code:

<?

include("common.php");



$qr1 = mysql_query("SELECT * FROM cats WHERE id=$id");

$cat = mysql_fetch_object($qr1);

$seo_title = $sitename." - ".$cat->name." Games";



include("header.php");

?>

<h2><?=$cat->name?> Games</h2>

<?

$qr1 = mysql_query("SELECT * FROM games WHERE cat=$id AND status=1");

while( $row = mysql_fetch_object($qr1) ){

?>

<table width=100%>

<tr>

<td align=left width=25%>

<a href="<?=$siteurl?>/games.php?id=<?=$row->id?>" class="nav"><img src="<?=$siteurl?>/uploads/thumb/<?=$row->thumb?>" border=0 width="70" height="59"></a>

</td>

<td align=left><div align=left>

<strong><a href="<?=$siteurl?>/games.php?id=<?=$row->id?>" class="nav"><?=stripslashes($row->name)?></a></strong><br>

<?=stripslashes($row->description)?><br>

<a href="<?=$siteurl?>/games.php?id=<?=$row->id?>" class="nav">Play</a>

</div></td>

</tr>

</table>

<br>

<?

}

?>

<?

include("footer.php");

?>

Anyone know? Im new to using db's in php so im pretty sure its something like that. Like itstrying to retrieve information from the database but its not responding. Anyone got any idea?

[End User]

Code:

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/xxvatarx/public_html/games/cats.php on line 5

99% of the time this is caused by no results being returned from the query. The error occurs because you're trying to do something with data that doesn't exist.

[xxvatarxx]

thnx, that helped alot more than you'd think, managed to fix it kinda i replaced the query with this

$sql = 'SELECT * FROM `cats` ORDER BY `id` ASC LIMIT 0, 30';

But now it is only turning up the first catagory, how do i get it to show up all the catagories?

[End User]

You may need to do something like this:

$sql = 'SELECT * FROM `cats` ORDER BY `id` WHERE 'id' <> '' ASC LIMIT 0, 30';