[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
[xsl] XPath & generate-id
Subject: [xsl] XPath & generate-id From: "Bhandari, Ashish" <Ashish.Bhandari@xxxxxx> Date: Mon, 31 Mar 2003 11:48:11 -0500 |
The named template below generates the following output - the href has no value when it should have one. <payerPartyReference href=""> </payerPartyReference> edcSystem is the root Element and the XPATH does point to a valid node in the XML tree. How do I specify an absolute path to generate-id function ?. I think it interprets the path as a relative one. Named Template: <xsl:template name = "eqs-utility:generate-pay-receive-references"> <xsl:param name = "payReceiveCode"/> <xsl:element name = "payerPartyReference"> <xsl:attribute name = "href"> <xsl:choose> <xsl:when test = "$payReceiveCode=0"> <xsl:value-of select = "generate-id(/edcSystem/EDCContract/@a10)"/> </xsl:when> <xsl:otherwise> <xsl:value-of select = "generate-id(/edcSystem/EDCContract/@a11)"/> </xsl:otherwise> </xsl:choose> </xsl:attribute> </xsl:element> </xsl:template> Thanks, Ashish. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl]Passing Parameter to the U, Leonidas Kanellos | Thread | RE: [xsl] XPath & generate-id, Michael Kay |
RE: [xsl] generate full xpath name , Michael Kay | Date | Re: [xsl] empty element as an end-m, Wendell Piez |
Month |