I have to following XML:
<root>
<a></a>
<b></b>
<a></a>
<a></a>
<b></b>
<c></c>
</root>
The order of a, b and c elements is random.
Now I want to sort the elements in a predefined way (first b, then a, then c).
I tried the following xslt:
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="@*">
<xsl:sort select="name()"/>
</xsl:apply-templates>
<xsl:apply-templates select="node()">
<xsl:sort select="name()"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Which sorts the element by name, thus a,b,c as expected.
Is there a way to define the order of the sort other then descending/ascending?
Thanks!
Now I want to sort the elements in a predefined way (first b, then a,
then c).
Here's one way:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes"/>
<!-- identity transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/root">
<xsl:copy>
<xsl:apply-templates select="b"/>
<xsl:apply-templates select="a"/>
<xsl:apply-templates select="c"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Here's another:
XSLT 2.0
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes"/>
<!-- identity transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/root">
<xsl:copy>
<xsl:apply-templates select="*">
<xsl:sort select="index-of(('b', 'a', 'c'), name())" />
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>