You say that the path expression points to a valid node in the XML tree.
The evidence suggests that it doesn't.
Perhaps you've fallen into the old trap of having a default namespace
declaration? Or perhaps the context node is in a different document from
the one you are assuming?
Show us the source file and the full stylesheet.
Michael Kay
Software AG
home: Michael(_dot_)H(_dot_)Kay(_at_)ntlworld(_dot_)com
work: Michael(_dot_)Kay(_at_)softwareag(_dot_)com
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Bhandari, Ashish
Sent: 31 March 2003 17:48
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] XPath & generate-id
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.
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