Display img sec using PHP

[GeorgeEhr]

I have a database where I store images, or rather paths to images.
I can display the images using http://www.domain.com/show.php?img=1.jpg&mid=1

Using that page it shows 4 images in a table 2 side by side and 2 side by side on the next row.

I want to call the show.php page from another file using the <img src>. Is it even possible?

[TheServant]

So I think that you're wanting to dynamically create an image of a page to show in another page using an <img /> tag?

This is possible using PHP GD Library, although I have never tried. I don't have much time at the moment to research this, but I have given you a few key words to google with. Also read a bit of what the GD Library can do here.

[Canabeez]

There's no need to use the GD library if you don't want to combine images. You can dynamically draw them on the page using either PHP of JS.

If you want to use PHP to output and image, then just:

<?php
    ob_start();
    $imageId = isset($_GET['img']) ? addslashes($_GET['img']) : false;
 
    if(!$imageId){
        exit;
    }
 
    $result = mysql_query("
        SELECT `image` FROM `images` WHERE `image_id` = {$imageId} LIMIT 1
    ");
 
    $imageObject = mysql_fetch_object($result);
    $image = $imageObject->image;
 
    header("Content-Type: " . mime_content_type($image) . ";\n");
 
    echo file_get_contents($image);
    ob_end_flush();
    exit;
?>

[Markus]

Canabeez, the output-buffering is redundant there, as you are not outputting anything before the call to header().

Mark (the Pedantic).

[Canabeez]

Thank's Mark, you completely right.