I've searched this and other forums and found what are clearly the right answers for some people but not for me.
I have a database with an item table. Item has a field called image that stores a path to an image. Here is an actual example of a path in case I've done something wrong:
without the quotes.
My code to display the image is:
$query = "SELECT * FROM item WHERE item_id = '1'"; $result = mysql_query($query); echo "<img src ='$result'>";
All I am gettin g returned to the browser is a broken image type square. When I check it with Firebug it tells me it is returning "Resource id # 5".
Can anone tell me in simple terms what I am doing wrong?