Subject: RE: [xsl] How to open a page in xsl From: Archana Rao <archana_heroor@xxxxxxxxx> Date: Wed, 24 Sep 2003 17:45:59 -0700 (PDT) |
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 > __________________________________ Do you Yahoo!? Yahoo! SiteBuilder - Free, easy-to-use web site design software http://sitebuilder.yahoo.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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