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
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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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