confilcting code ? Please advise

Discussion in 'PHP' started by Funk-woo10, Dec 18, 2007.

  1. #1
    Hi

    On a page of my site i am running a couple of different sql queries.

    I seem to have a problem though, I have wriiten some new code that seems to conflict with the code already on the page.

    Error message -

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


    Sample of code: (Code already on the site)

    function getRating($photoid)
    {
    	$rating="Not Rated yet!";
    	$sql="select * from ratings where photosid=$photoid";
    	$res=mysql_query($sql);
    	while($obj=mysql_fetch_object($res))
    	{
    		$rating="Rated $obj->avgrating from $obj->numvotes votes";
    	}
    	return $rating;
    PHP:

    Sample of code2: (Code I am trying to develop, works fine until put on the same page as the code above.)

    <?php
    
    
    $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";
    $result=mysql_query($sql);
    while($object = mysql_fetch_object($result))
       {
    
       	echo "<td>\n";
    
       	if($object->url=="")
       	{
       		$img="<a href='http://****/index.php?&username=$object->username&phid=$object->photosid'><img border=0 width=50 height=50 src='../pics/$object->filename'></a>";
       	}
       	else
       	{
       		$img="<a target='_blank' href='$object->url'><img  border=0 width=50 height=50 src='$object->url'></a>";
       	}
    
       	$idt=$object->photosid;
       	$username=$object->username;
       	print $img;
       	?>
           	<br /><a href="http://****/index.php?cmd=20&username=<? print $username; ?>" class='footer'>View <? print $username; ?>'s profile</a>
           <?php
    
       	echo "</td>";
    
    
       	{
       		echo "\n\n";
       	}
       }
    
       ?>
    PHP:
     
    Funk-woo10, Dec 18, 2007 IP
  2. AsHinE

    AsHinE Well-Known Member

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    #2
    It can be that query is errorneous so mysql does not return a valid result.
    Try mysql_error to show errors.
     
    AsHinE, Dec 18, 2007 IP