I thinnk you are confused between what the XSLT stylesheet does and what
the generated HTML does. It's the generated HTML that does things like
"redirecting to different pages". Do you know what HTML you want your
stylesheet to generate? If so, we can help. If not, you're asking the
wrong list.
Michael Kay
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Archana Rao
Sent: 25 September 2003 01:46
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] 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(_at_)yahoo(_dot_)com"/>
<xsl:template match="/">
<xsl:if test="contains($userid, 'yahoo')">
<!-- Goto http://www.yahoo.com -- >
</xsl:if>
</xsl:template>
</xsl:stylesheet>
Hope this helps,
Basically i want this xsl to redirect to different
pages depending on the domain name in the userid.
Thanks,
Archana
--- Michael Kay <mhk(_at_)mhk(_dot_)me(_dot_)uk> wrote:
So i know what the username is, now my problem is
i am
trying to open up www.yahoo.com if the username is
xyz(_at_)yahoo(_dot_)com and www.hotmail.com if the username
has xyz(_at_)hotmail(_dot_)com(_dot_)
I know i can use <xsl:if test="contain($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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