Re: I'm really stuck!

Hi Jennifer. Change {$session["uid"]} to {$session['uid']} --> double quotes to single quotes. It's inside a set of single quotes already, so you have to use double quotes. /bsh/ Jennifer Downey wrote:

Hi all, I'm really stuck and I'm not asking anyone to re-write this just show me what is wrong, or explain why it wont work. It just seems logical that this should work. The first query will print the pet id in the browser. $query="SELECT id FROM wt_users WHERE uid={$session["uid"]}"; $ret = mysql_query($query); while(list($pet)= mysql_fetch_row($ret)) print("<BR>your pet id is $pet"); So if $pet will print the id from the wt_users (has a value of 3) and I assign $id = $pet (id also has a value of 3 in image_data) why doesn't it show the image? if($id) { $id = $pet; $query = "select bin_data,filetype from image_data where id=$id";

     $result = mysql_query($query);

$data = mysql_query($result,0,"bin_data"); $type = mysql_query($result,0,"filetype"); Header( "Content-type: $type"); echo $data; }; echo "<img src=\"petdata.php?id=$id\">"; Thanks Jennifer Downey

-- /---------------------------------------------=[ BILLY S HALSEY ]=--\ | Member of Technical Staff - Sun Microsystems, Inc. | | ESP Solaris Software - Software Problem Resolution Team | | "No day but today." - J. Larson, RENT x55403/858-526-9403 | \--=[ bsh@sun.com ]=------------------------------------------------/