You send way too much info, forcinf people to read through all of it, to
understand your way of thinking and your problem.
Keep it simple.
Anyway, I think there is something wrong in here
$sql = "SELECT furn_pic";
$sql .= "FROM crost";
$sql .= "WHERE item_num = $item_num";
$result = mysql_q
I am trying to display a gif file that I've
successfully stored up to mysql. I have 3 short files
that all work to gether to accomplish this task.
1. get_image.html (no problems w/this one)
2. get_image.php (I think this one is fine too)
3. crostchair.php (this one has the problem)
It is w
