Some suggestions...

On Monday, December 30, 2002, at 03:55  PM, Nikos Gatsis wrote:
where showpict.php:

 $query="SELECT pict FROM pict WHERE pro_id= '$pro_id";
(You're missing an end ' there, but apparently that's not the problem)

$result=mysql_db_query($database, $query, $conn) or Die (mysql_error());
$type = $photo_type;
Where did the $photo_type variable come from? What's in it?

 if (!empty($photo)) {
  header("Content-Type: {$type}");
I believe this header should be in the form header("Content-Type: image/jpeg") or whatever image type you have.

You should probably include an additional header
header("Content-Length: " . strlen($photo));

  echo $photo;


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