[RESOLVED] write a page with multiple jpegs

clairence

I pretend to know php, but I am constantly flummoxed by what turns out to be very simple stuff. So here's my current quandry.

I have a script that pulls an image from a protected directory and displays it as a new image page.

It turns out that some of the pages I want to display will have more than one image. I can't figure out how to write more than one jpeg to a page.

if($images){    

		 header('Content-Type: image/jpg'); 
	  	 foreach($images as $ikey=>$ival){
			$page='jpegdirectory/'.$ival;
			$display=imagecreatefromjpeg($page);	 
			imagejpeg($display); 
		 }
	  }

This only displays the first jpeg and stops. How to display all jpegs in a page?

Thanks.

bradgrafelman

The problem is what you're referring to as a "page" doesn't exist. What your script is doing is outputting the binary data of a file. You can't simply concatenate the binary data of a bunch of files and expect it to mean anything.

If you want to display multiple images, then you'll need an intermediate PHP script that outputs an HTML page with multiple <img> tags. Each <img> tag could then point to a script containing code much like what you have above, only without any loops (its job would be to output a single image, e.g. one whose filename or some other unique identifier is passed to it in the URL).

EDIT: Also note that there's no reason to go through the overhead of using the GD library to load the image and then re-render it; just use [man]readfile/man to copy the binary data from the file directly to the output.

sneakyimp

yes you can't just loop through a bunch of images and output their contents from you script because the browser that requested you php page will think that all of the images you are spitting out are one big jpeg file. it sounds to me like your browser might be downloading all the images and only displaying the first one.

you would need to either alter your script to create a new, big jpeg and place the other jpegs in this new image and then output that (which would be a real coding chore and would tax your server for no reason) or you can just output an HTML page which has multiple IMG tags, each of which refers to your script and supplies it with some identifier that can be used to locate the correct image.

show this to the visitor:

<html>
  <body>
    <img src="my_php_script.php?q=x">
    <img src="my_php_script.php?q=y">
    <img src="my_php_script.php?q=z">
  </body>
</html>

and then in my_php_script.php, you would put something like this:

$image_key = $_GET['q'];

// code here to read the image from the protected folder and just readfile it like bg says.

clairence

Thank you both for pointing me in the right direction.