I kind of understand your problem so I will take a stab at it.
First of all, a clarification. If you pass a url that ends with something like: fader.js?r=asp&r=cop&r=ste - you are not passing the values of the variables asp, cop, and ste. As it stands you are passing three different values for "r." If you were to pass values for the variables asp, cop and ste, the url would look like this: fader.js?asp=aspvalue&cop=copvalue&ste=stevalue
Secondly, if you just want to pass those values from page1 to page2 via a url, you don't need an array on page2 to do so. In PHP, those variables passed via the url are automatically available as variables on page2. Just create a test.php document that contains the code <?php echo $variables; ?> and view it with the url http://locationoffile/test.php?variables=whatever
to see. So if you are passing fader.js?asp=aspvalue&cop=copvalue&ste=stevalue to a page, you can just go right into the db code and say $sql = "SELECT * FROM table WHERE asp_column=$asp AND cop_column=$cop AND ste_column=$ste" and it will work.
If you really want to create an array, there are a couple of ways. I never use it but I think those variables are available in some array of global scope. The second, and the way I would do it is to use the parse_url() function in PHP. It breaks a url into constituent parts and returns an array. You can then explode the "query" part of the array into another array like so:
$parts = parse_url(genenv('REQUEST_URI'));
$query = explode("&", $parts[query]);
$query now contains an array of all the variable=value pairs. You can just explode() them out any way you want.
I hope somethign here answers your question!
- Erik