How to open a page in xsl

Hello everybody,

JSP passes the username to my xsl, i know what the
user name is that is, in my xsl i have <xsl:param
name="username"/>

So i know what the username is, now my problem is i am
trying to open up www.yahoo.com if the username is
xyz@xxxxxxxxx and www.hotmail.com if the username has
xyz@xxxxxxxxxxxx

I know i can use <xsl:if test="contain($username,
'yahoo')"> to check for the username, but then i don't
know how to specify in the <xsl:if> to open up
www.yahoo.com.

Hope you understood my problem.

TIA,
Archana

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