Question

I have media_main.php that is obtaining data from a MySQL database. One of the pieces of data obtained is “filename” using the mySQL SELECT statement. On media_main.php, there is a link to open player.php which has a flash MP3 player embedded. The link to open the new window is as follows:

<a href="javascript:;" onclick="return popitup('player.php')" title="Listen">

I would like “filename” to be passed from media_main.php to the player.php page. I think I need to use GET or POST, but am not sure how to since the data isn’t coming directly from a form.

Answer 1

You can achieve this by reading $_GET['file'] on player.php, and setting it on media_main.php.

On media_main.php you would use PHP to make the link dynamic:

<a href="javascript:;" onclick="return popitup('player.php?file=<?php echo $Data['filename']; ?>')" title="Listen">

Which would result in:

<a href="javascript:;" onclick="return popitup('player.php?file=dynamic_file.mp3')" title="Listen">

Then on player.php, you would echo $_GET['file'] where you want the file to be included.

Also, be sure to validate all data. To make it safe to display on your website, you can wrap $_GET['file'] with htmlentities(), so that a user can’t use the $_GET['file'] to display HTML on your site.