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


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