PHP Image script

Question

So my plan is instead of adding a style to each picture or making different images with different sizes, I made this script:

/*FILENAME image.php*/
    if(isset($_GET['size'])){
        $image['size']  = $_GET['size'];
        $size = explode('x', $image['size']);
        $image['width'] = $size[0];
        $image['height'] = $size[1];
    } else {
        $image['size']  = null;
    }
echo "<img src='images/".$image['name'].".jpeg' style='width:".$image['width'].";   height:".$image['height'].";'>";

But it doesn't work, In one of my scripts, I have this:

echo '<img id="img_01" src="image.php?name='.$row2['name'].'&size=150x160" data-zoom-image="image.php?name='.$row2['name'].'&size=150x160"/>';

And the browser echo's just a missing image. How could I make my script work? I know there must be a way because I remember doing something like this with an open source.

chasepeeler · Accepted Answer · 2014-01-18 05:05:27Z

1

That's not how dynamic images work. Image.php has to actually output the raw binary content of an image (open a jpg file in a text editor, that's what the PHP file should generate).

Take a look at the gd functions, like http://www.php.net/manual/en/function.imagecopyresized.php, which should have some examples that will help

answered Jan 18, 2014 at 5:05

chasepeeler

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