Reputation: 1022
I want to sort parent node, based on child's name attribute.
XML unsorted:
<grand-parent>
<parent>
<child name="c"/>
<child_next name="a"/>
</parent>
<parent>
<child name="a"/>
<child_next name="a"/>
</parent>
<parent>
<child name="b"/>
<child_next name="a"/>
</parent>
</grand-parent>
Expected Output:
<grand-parent>
<parent>
<child name="a"/>
<child_next name="a"/>
</parent>
<parent>
<child name="b"/>
<child_next name="a"/>
</parent>
<parent>
<child name="c"/>
<child_next name="a"/>
</parent>
</grand-parent>
XSLT In Use (not working - returns the same XML that is input):
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes" method="xml" encoding="UTF-8"/>
<xsl:strip-space elements="*"/>
<xsl:template match="node()|@*" name="identity">
<xsl:copy>
<xsl:apply-templates select="node()|@*">
<xsl:sort select="/grand-parent/parent/child/@name" order="descending" />
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Upvotes: 1
Views: 4428
Reputation: 52858
I would take the sort out of the identity transform:
XSLT 1.0
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="grand-parent">
<xsl:copy>
<xsl:apply-templates select="parent|@*">
<xsl:sort select="child/@name" data-type="text"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Output
<grand-parent>
<parent>
<child name="a"/>
<child_next name="a"/>
</parent>
<parent>
<child name="b"/>
<child_next name="a"/>
</parent>
<parent>
<child name="c"/>
<child_next name="a"/>
</parent>
</grand-parent>
Upvotes: 4