RE: How to open a page in xsl

Herez the piece of code.

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:param name="userid" select="xyz@xxxxxxxxx"/>

<xsl:template match="/">                              
                         
  <xsl:if test="contains($userid, 'yahoo')">
     <!-- Goto http://www.yahoo.com -- >
  </xsl:if>    
</xsl:template>

</xsl:stylesheet>

Hope this helps, 

Basically i want this xsl to redirect to different
pages depending on the domain name in the userid.

Thanks,
Archana

--- Michael Kay <mhk@xxxxxxxxx> wrote:
> > 
> > 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.
> > 
> 
> No, sorry, I don't. I haven't the faintest idea what
> you mean by
> "opening up www.yahoo.com". XSLT transforms a source
> tree into a result
> tree, where does "opening up" a web site fit into
> this?
> 
> Michael Kay
> 
> 
>  XSL-List info and archive: 
> http://www.mulberrytech.com/xsl/xsl-list
> 


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 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list