Displaying an image in php using mysql - php

Question

I'm trying to display an image along with other information from the database.
PHP
<?php
mysql_connect("localhost","111","222") or die("Could not connect to localhost");
mysql_select_db("movies") or die( "Could not connect to database");
$result = mysql_query("SELECT * FROM allmovies");
if ($result == false ) die(mysql_error());
while($row = mysql_fetch_array($result))
{
echo "<img src=' . $row['image'] . '>";
?>
Like:
Title: Blah
Price: Blah
Image: <img src=rarara">
All from MySQL in one page?

Answer 1

Don't store image data in a database, they are generally not suited to this and incurs extra overhead on your MySQL connections returning the data. You should be storing the path to the image file, and serving that.
If you insist on doing it you you should only be returning one image at a time with the proper header based on the image type using something like the following:
$imagedata = data_from_mysql();
header('Content-Length: ' . sizeof($imagedata) );
header('Content-Type: image/png');
echo $imagedata;
exit;
If you really want to make your page source bloated, slow, unmanageable, and nigh-uncacheable:
while( $imagedata = data_from_mysql() ) {
echo "<img src='data:image/png;base64," . base64_encode($imagedata) . "'>";
}
I cannot stress enough how these are terrible ideas that you should not use, but if you cannot listen to reason you can at least do bad things the right way.

Answer 2

You could use imagecreatefromstring()
$im = imagecreatefromstring($row['image']);
if ($im !== false) {
ob_start();
imagejpeg($im);
$data = ob_get_contents();
ob_end_clean();
echo '<img src="data:image/jpg;base64,' . base64_encode($data) . '" />';
}
Just my opinion, but it might be slightly more sane to save the images to the file server and then store a reference to the path instead of the whole image as a blob?