Passing an Variable from one PHP File to another - php

Question

I'm working on an image resizer, to create thumbnails for my page. The resizer works on principle of include a DIRECT link to the image. But what I want to do is put in the PHP Variable in the URL string, so that it points to that file and resizes it accordingly.
My code is as follows :
<img src="thumbnail.php?image=<?php echo $row_select_property['image_url']; ?>
Image Resize :
<?php
// Resize Image To A Thumbnail
// The file you are resizing
$image = '$_GET[image_url]';
//This will set our output to 45% of the original size
$size = 0.45;
// This sets it to a .jpg, but you can change this to png or gif
header('Content-type: image/jpeg');
// Setting the resize parameters
list($width, $height) = getimagesize($image);
$modwidth = $width * $size;
$modheight = $height * $size;
// Creating the Canvas
$tn= imagecreatetruecolor($modwidth, $modheight);
$source = imagecreatefromjpeg($image);
// Resizing our image to fit the canvas
imagecopyresized($tn, $source, 0, 0, 0, 0, $modwidth, $modheight, $width, $height);
// Outputs a jpg image, you could change this to gif or png if needed
imagejpeg($tn);
?>
What I am trying to do is pass on the variable "image=" to the Thumbnail script. At the moment I am passing it through the URL string, but it doesnt seem to load the graphic.
I'll try expand on this more, should you have questions as I am finding it a little difficult to explain.
Thanks in advance.

Answer 1

I suspect at least part of the problem is that your existing...
$image = '$_GET[image_url]';
...line is creating a text string, rather than getting the contents of the 'image_url' query string. Additionally, your passing in the image name as "?image=" in the query string, so you should simply use "image", not "image_url".
As such, changing this to...
$image = $_GET['image'];
...should at least move things along.

Answer 2

Change it
$image = '$_GET[image_url]';
to
$image = $_GET['image'];

Answer 3

$image = '$_GET[image_url]';
should be
$image = $_GET['image'];