RE: [xsl] How to open a page in xsl

["Jim Fuller" <jim.fuller@xxxxxxxxxxxxxxxxxx>]

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@xxxxxxxxx]
Sent: 22 September 2003 22:41
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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@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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