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, $conn) or Die (mysql_error());
list($photo)=mysql_fetch_row($result);
$type = $photo_type;
Where did the $photo_type variable come from? What's in it?
I believe this header should be in the form header("Content-Type: image/jpeg") or whatever image type you have.if (!empty($photo)) { header("Content-Type: {$type}");
You should probably include an additional header
header("Content-Length: " . strlen($photo));
echo $photo; } THANX Nikos
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