Subject: [xsl] Problem in making choices using <xsl:choose> From: "Sreekanth Pallavoor" <sreekanth@xxxxxxxxx> Date: Thu, 31 May 2001 19:37:26 -0700 |
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> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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