将变量从一个PHP文件传递到另一个

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

$image = '$_GET[image_url]';

应该

$image = $_GET['image'];

Answer 3

Change it

 $image = '$_GET[image_url]'; 

to

 $image = $_GET['image'];