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#1 AncientSage

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Posted 08 August 2006 - 09:22 PM

Hello,

If I create an image, using this script:

function dbar($status) {
   $bar = ImageCreate(100, 5);

   $red = ImageColorAllocate($bar, 255, 0, 0);
   $green = ImageColorAllocate($bar, 0, 255, 0); 

   ImageFilledRectangle($bar, 0, 0, $status, 4, $green);
   
   header("Content-type: image/png");
   ImagePng($bar);

   ImageDestroy($bar);
}
$status = '10'; //Ignore the status var...
dbar($status);

Then, I go and save it to a php file, is it possible, instead of saving it into a seperate file, and then including it into an <img> tag, that I could have it included via an HTTP var? Like...

<img src="script.php?img=phpimage">

Instead of...

<img src="image.php">

That way, I am able to get the $status var (as it will be defined by a database, probably). But the image is only displayed if included in an <img> tag as script.php?img=phpimage.

Thanks.

#2 corbin

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Posted 08 August 2006 - 09:58 PM

call to it like <img src="script.php?image=y"> and then you could do
[code=php:0]
if($_GET['image'] == "y") {
function dbar($status) {
  $bar = ImageCreate(100, 5);

  $red = ImageColorAllocate($bar, 255, 0, 0);
  $green = ImageColorAllocate($bar, 0, 255, 0);

  ImageFilledRectangle($bar, 0, 0, $status, 4, $green);
 
  header("Content-type: image/png");
  ImagePng($bar);

  ImageDestroy($bar);
}
$status = '10'; //Ignore the status var...
dbar($status);
}
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#3 AncientSage

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Posted 09 August 2006 - 02:07 AM

Thanks, that works.

I wasn't sure if GET vars could be passed via an image.

However, rather than make it call to an undefined function, I put the check around the function call, and let the function be defined.




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