Steve a. its a typo of mine in the e-mail b. the "_type" is the type of the binary data. From the PHP manual read: "$userfile_type - The mime type of the file if the browser provided this information. An example would be "image/gif" c. I'll try the file size
Thank you anyway ----- Original Message ----- From: Steve Cayford To: Nikos Gatsis Cc: PHP-mailist Sent: Tuesday, December 31, 2002 1:07 AM Subject: Re: [PHP-DB] SHOW PICTURE FROM DATABASE 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 $photo_type variable come from? What's in it? > if (!empty($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; > } > > THANX > Nikos -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php