Extract Picture From Database?

Volitics

Can anyone tell me how to extract an image from a MySQL database then display the image on a page?

The image is stored in binary format in the MySQL database. I uploaded the image with the phpMyAdmin program.

I have got just a standard SELECT query like the one below:

while ($Row = mysql_fetch_array($Result)){
$Image = "".$Row["Picture1"]."";
echo "<img src=\"$Image\">";
}

However, when I run the script instead of outputting the image on the page all I get is a real long string of characters like this:
"ÍEñ5ñs›ä¨£¥¯+¦õmY=/{yXö[Òµí6=KG¾†öÑþì‘6FG" etc. etc.

I would appreciate any help.

Thanks.

superwormy

You can't do what you're trying to do because you're trying to echo out binary data on a web page.

Images are seperate things, and seperate requests are made by the browser to get those images.

Instead, you should echo the filepath to the image:
<img src="/path/to/php/file_which_makes_a_query_sends_a_header_and_outputs_image.php?getImageFromDB=imageFromDB.jpg">

And then this file:
path/to/php/file_which_makes_a_query_sends_a_header_and_outputs_image.php

Would make a query for $_GET['getImageFromDB'], send a header("Content-type: image/jpeg"); ( or PNG/GIF/etc. ) and then print the binary data.

I should note that the MySQL manual specifically reccomends AGAINST storing images in a database because of performance reasons.

Volitics

superwormy;

Thank you for responding to my problem.

I've been reading other postings in this forum and in the DevShed forum.

It seems like a lot of folks prefer using a file folder storage system rather than storing images in a database.

I've already figured out how to upload images and store them in a file folder. Considering all of the difficulties with database storage maybe that would be the best way to approach the problem.

Thanks again.

Volitics

I can't get the script below to work. Can anybody see what I'm doing wrong?

// ------ Image page.
echo "<img src=\"ImageOutput.php?PictureDB=PictureJPEG_98x98_10.jpg\">";

// ----- ImageOutput.php
require_once "../conf/config.inc.php";
$Query = "SELECT * FROM RealtorHouses where filename = 'PictureJPEG_98x98_10.jpg'"; 
$Result = mysql_db_query ($db_info[dbname], $Query, $db_connection);
while ($Row = mysql_fetch_array($Result)){

Header( "Content-type: image/jpeg");
$PictureDB = $_GET['filename'];
	}

Thanks.