XML - XSLT Parameter Question

[willyp]

Hello,

I am developing a website that has a drop down menu populated with 'user types' ie: customers, buyers, ect...

And below this menu is a list of links.

This list of links changes based on the selection of user type chosen.

So the XML would look like this for each link:

<link>
<title>Headline </title>
<url>link-to-page.html</url>
<users>customer s</users> <!--User type(s) in this element-->
</link>

So far we have been filtering our data in the following way through XSL:

<xsl:if test="contains( users,'customer s')">


This works fine, however, we would like a way of passing a variable to
the xsl file so that we don't have to have a different xsl file to
filter each user role.

So perhaps it would look like this:

<xsl:if test="contains( users,'$user')" >

with

$user

being the variable.

Is it possible to pass a variable from the browser (via javascript or
something else) to the xsl file so that we can use one xsl file as our
filter?

Thanks a lot, and if this was completely confusing, I apologize.

-WillyP

[Dököll]

Greetings!

Have a look here:

http://www.w3schools.com/xlink/default.asp