Subject: Re: [xsl] returning nodes (not a string) using variable-choose. From: "Dimitre Novatchev" <dnovatchev@xxxxxxxxx> Date: Fri, 23 Jun 2006 15:39:16 -0700 |
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.
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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