a sample src xml and desired xml may reveal what you want to do, possibly
you want a choose statement
<xsl:choose>
<xsl:when test="contain($username,'yahoo')">
do something
</xsl:when>
<xsl:otherwise>
do something by default
</xsl:otherwise>
</xsl:choose>
not sure at all what you want to do ?
gl, jim fuller
-----Original Message-----
From: Archana Rao [mailto:archana_heroor(_at_)yahoo(_dot_)com]
Sent: 22 September 2003 22:41
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] How to open a page in xsl
Hello everybody,
JSP passes the username to my xsl, i know what the
user name is that is, in my xsl i have <xsl:param
name="username"/>
So i know what the username is, now my problem is i am
trying to open up www.yahoo.com if the username is
xyz(_at_)yahoo(_dot_)com and www.hotmail.com if the username has
xyz(_at_)hotmail(_dot_)com(_dot_)
I know i can use <xsl:if test="contain($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.
TIA,
Archana
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