i would like to get distinct nodes from my xml on multiple levels. Can anyone please give me some hints how to do this? The methods i googled (Muenchian method, for-each-group) were explained with single grouping keys and plain hierarchy.
Here's an example of my xml:
<persons>
<person>
<name>Tom</name>
<age>20</age>
<mails>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
</mails>
</person>
<person>
<name>Tom</name>
<age>20</age>
<mails>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
</mails>
</person>
</persons>
I would like to have distinct person nodes based on name and age, and also a distinct set of mail-nodes. So for the example the desired output would be:
<persons>
<person>
<name>Tom</name>
<age>20</age>
<mails>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
</mails>
</person>
</persons>
Is there a way to do this? Thanks a lot in advance.
I. XSLT 1.0 solution:
This transformation:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="kPersByNameAndAge" match="person"
use="concat(name, '+', age)"/>
<xsl:key name="kmailByNameAndAge" match="mail"
use="concat(../../name, '+', ../../age)"/>
<xsl:key name="kmailByNameAndAgeAndVal" match="mail"
use="concat(../../name, '+', ../../age, '+', .)"/>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/*">
<persons>
<xsl:apply-templates select=
"person[generate-id()
=
generate-id(key('kPersByNameAndAge',
concat(name, '+', age)
)
[1]
)
]
"/>
</persons>
</xsl:template>
<xsl:template match="mails">
<mails>
<xsl:apply-templates select=
"key('kmailByNameAndAge', concat(../name, '+', ../age))
[generate-id()
=
generate-id(key('kmailByNameAndAgeAndVal',
concat(../../name, '+', ../../age, '+', .)
)
[1]
)
]
"/>
</mails>
</xsl:template>
</xsl:stylesheet>
when applied on the provided XML document:
<persons>
<person>
<name>Tom</name>
<age>20</age>
<mails>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
</mails>
</person>
<person>
<name>Tom</name>
<age>20</age>
<mails>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
</mails>
</person>
</persons>
produces the wanted, correct result:
<persons>
<person>
<name>Tom</name>
<age>20</age>
<mails>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
<mail>[email protected]</mail>
</mails>
</person>
</persons>
II. XSLT 2.0 solution
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="kmailByNameAndAge" match="mail"
use="concat(../../name, '+', ../../age)"/>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/*">
<persons>
<xsl:for-each-group select="person" group-by="concat(name, '+', age)">
<xsl:apply-templates select="."/>
</xsl:for-each-group>
</persons>
</xsl:template>
<xsl:template match="mails">
<mails>
<xsl:for-each-group select=
"key('kmailByNameAndAge', concat(../name, '+', ../age))"
group-by="concat(../../name, '+', ../../age, '+', .)"
>
<xsl:apply-templates select="."/>
</xsl:for-each-group>
</mails>
</xsl:template>
</xsl:stylesheet>
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