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Answer 1

A:

Looks like you're trying to include a javascript file with PHP.. echo ' instead..

Prot0 2010-08-05 12:49:25

echo '../' . $PAGE_URL; gives me <br/><br/>Unexpexted output: ../upload_application.php, referer: http://10.0.0.2/jtpc/index.php

shay 2010-08-05 12:55:23

Answer 2

A:

Hi, it looks like what you're trying to do is include a javascript file on demand. Without the full source (including what you're passing in the URL, it's a bit difficult to debug.

I feel like what you want to be saying here:

$response->fields->frame_main = file_get_contents('../' . $PAGE_URL);

is actually

$response->fields->frame_main = '../' . $PAGE_URL;

Since based on the output you've provided, it looks like you're accidentally reading the file into the tag, rather than providing the filename.

One small point I feel obligated to bring up. Your code in it's current implementation seems like it would allow me to pass a POST variable like "../../../../../etc/passwd" and retreive the contents of that file. If you do end up reading in files on demand from the user, please validate them fully. Filtering them using a function like basename() may be of use.

preinheimer 2010-08-05 15:16:29

how come they are the same ? , frame_main ,is the id of the div , my ajax will replace the div.innerHTML with that content

shay 2010-08-05 15:51:45

this is the way i got it working , function getPageContent( ob_start(); require_once( '../' . $PAGE_URL ); $contents = ob_get_contents(); ob_end_clean(); $response->fields->frame_main = $contents;} but i had to separate the javascript file ,can i do that without separating the js ?

shay 2010-08-05 16:20:52

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php include gives an error

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