Subject: Re: returning nodes (not a string) using variable-choose.
From: "Dimitre Novatchev" <dnovatchev@xxxxxxxxx>
Date: Fri, 23 Jun 2006 15:39:16 -0700
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Read about and understand the difference between xsl:value-of and xsl:copy-of.
Also, try to avoid creating an RTF which then must be converted it
into a nodeset using the xxx:node-set() extension function.
--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
On 6/23/06, Steve <subsume@xxxxxxxxx> wrote:
The below somewhat elaborate lines assign a value to $typeInfo,
however they don't return results as a nodelist, just a big string.
------
<xsl:variable name="typeInfo">
<xsl:choose>
<xsl:when test="$type='Other'">
<xsl:value-of select="$desc" />
</xsl:when>
<xsl:otherwise>
<xsl:variable name="table">
<xsl:choose>
<xsl:when test="$type='Organizational Contact'">
<xsl:text>CA_Orgs</xsl:text>
</xsl:when>
<xsl:when test="$type='Meeting or Training'">
<xsl:text>CA_Meetings</xsl:text>
</xsl:when>
<xsl:when test="$type='Travel'">
<xsl:text>CA_Travel</xsl:text>
</xsl:when>
</xsl:choose>
</xsl:variable>
<xsl:value-of
select="document(concat(concat(concat('http://server.org/getXML.asp?key=id&val=',$desc),'&table='),$table))"
/>
</xsl:otherwise>
</xsl:choose>
</xsl:variable>
<xsl:value-of select="$typeInfo" />
-----------------
Anyway to do this with XSLT 1.0 or am I pushing it?
--Steve
--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
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