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Problem in making choices using <xsl:choose>
- To: XSL-List at lists dot mulberrytech dot com
- Subject: [xsl] Problem in making choices using <xsl:choose>
- From: "Sreekanth Pallavoor" <sreekanth at tibco dot com>
- Date: Thu, 31 May 2001 19:37:26 -0700
- Reply-To: xsl-list at lists dot mulberrytech dot com
Hi,
I am new to XSLT. I am writing one XSLT program, where I need to print
different output depending upon different conditions. But the code which
checks for different conditions, doesn't seem to be working properly.
The relevant part of code is given below. Here I have got a variable
"state" getting set to some state code which comes from some external
source. If it's value is AZ, I want to print "Arizona" and if it's value
is anything else then I want to print "Other". The result I am getting
is, it always prints "Arizona" even if $state is set to something other
than "AZ". (To double-check, I am printing the $state value also in the
output. Even if it is CA, the output I still get is "CA Arizona" and not
"CA Other").
Can someone tell me what could be the reason.
Thanks.
-- Sreekanth
<xsl:template match="headline">
<xsl:choose>
<xsl:when test="contains($state,'AZ')">
<font face="{$headline-font}"
size="{$headline-size}"><xsl:value-of select="$state"/> Arizona </font>
</xsl:when>
<xsl:otherwise>
<font face="{$headline-font}"
size="{$headline-size}"><xsl:value-of select="$state"/> Other </font>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
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