Hello and a BIG Thank-You!!! To you who have helped me...

This I think will be my last image question. Okay the user uploaded a picture , it went to the (Photos directory) (Name and Type Show) Great!... Next to the (Mysql Database under aphoto_filename) Again with (Name and Type Showing) Excellent!, It is in the users Mysql Database listing with all their other information. So Perfect! Last it shows up Beautiful!

Brick Wall I'm exhausted even missed school trying to solve it, I cannot do that again.
PROBLEM TO BE SOLVED:

When the user leaves the photo view page, I cannot get their photo to stay reappear. It remains in the photos directory and in the mysql database... When I try calling it... It only displays the image name and type I need it to open.

Please See My Code Below and Thank A Zillion!


<?
include'db.php';

$sql_check = mysql_query("SELECT aphoto_filename FROM gallery_photos WHERE email_address='".($email_address)."'AND password='".($password)."'" );

$aphoto_filename = mysql_fetch_array($sql_check,
MYSQL_ASSOC); //MYSQL_ASSOC;
$sql_check_num = mysql_num_rows($sql_check);
if($sql_check_num == 0){

if ($type == "image/pjpeg") {
imagejpeg($img);
} else if ($type == "image/x-png") {
imagepng($img);
} else if ($type == "image/gif") {
imagegif($img);
}
$handle = fopen($aphoto_filename['tmp_name'], $images_dir."/$filename". "wb");

echo ("");
} else {
echo $aphoto_filename['aphoto_filename'];
}

?>

P.S I actually have... aphoto, bphoto, cphoto and dphoto ... When anyone wants to see the users photo I need the photos to open. Thanks again soooooo much, puddin

Your overthinking this.

All you have to do is echo the HTML to display the image from the directory it's in.

[php]echo "<img src='/path/to/" . $filename . " ' />[/php]

No need to use fopen to open the image.

Gee Wiz

I must not be filling it in correctly...
echo "<img src='/path/to/" . $filename . " ' />

Is the path/to $images_dir And The $filename Is $aphoto_filename?
It's not coming up... I'll keep trying...