RE: Problem in making choices using <xsl:choose>

Sreekanth

Try <xsl:when test="$state='AZ'"> for your choose test, like so,

<xsl:template match="headline">
  <xsl:choose>
    <xsl:when test="$state='AZ')">
      <font face="{$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>

Hope that helps,

Tim Watts

-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Sreekanth
Pallavoor
Sent: Friday, 1 June 2001 12:37 PM
To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx
Subject:  Problem in making choices using <xsl:choose>

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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