I think this is a sql problem, but I can't seem to get rid of this error message

Answer 1

Try this (Without the ticks)

$sql = "SELECT * FROM images WHERE iimageid = '" . $iimageid . "'";

or (typecasting to integer)

$sql = "SELECT * FROM images WHERE iimageid = ". (int)$iimageid;

If its an integer value always typecast to integer, then any text string it will automatically set to 0 "For security purposes"

Answer 2

It looks like your SQL statement is invalid to me, try:

"SELECT * FROM images WHERE iimageid = '" . $iimageid . "'";

Answer 3

Ok nm, I found it pretty quickly after submitting this thing. $iimageid wasn't being pulled, taken care of now. :)

Answer 4

Looks like a PHP error to me, not MySQL. Those error codes are normally by the PHP interpreter finding something it doesn't like in your code. You also appear to have not closed off the <a> tag? And there appears to be an extra " after your 'thumb/' part of the <img> tag.

I would re-write this section;

        <?  while($display_info = mysql_fetch_array($runSQL)) { ?>
                    <a href="sfd/pimages/<? echo $display_info['vimage']; ?>" rel="lightbox[g]"><img src="sfd/pimages/thumb/"<? echo $display_info['vimage']; ?>">
                <br>
        <?  } ?>

As follows, see if that helps.

        <?  
        while ( $display_info = mysql_fetch_array($runSQL) ) { 
             print "<a href=\"sfd/pimages/{$display_info['vimage']}\" rel=\"lightbox[g]\">";
             print "<img src=\"sfd/pimages/thumb/{$display_info['vimage']}\">";
             print "</a><br>\n";
        } 
        ?>

Answer 5

Everytime someone uses mysql_* functions, baby raptor jesus eats a lolcat. You should start using PDO ( http://fr.php.net/manual/en/book.pdo.php ), and for the part mixing html and php you may prefer the alternative syntax http://www.php.net/manual/en/control-structures.alternative-syntax.php