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358319
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2008-12-11T02:50:32.323
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2008-12-11T02:50:32.323
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This question requires a little bit more detailed answer than just pointing to a good <a href="http://dpawson.co.uk/xsl/sect2/N4486.html#d6060e85" rel="nofollow noreferrer">Muenchian Grouping</a> source. The reason is that the needed grouping requires to identify both the names of all children of an "ele[SomeString]" element and their parent. <a href="http://dpawson.co.uk/xsl/sect2/N4486.html#d6060e238" rel="nofollow noreferrer">Such grouping</a> requires to define a key that is uniquely defined by both unique sources, usually via concatenation. This transformation: <pre> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:key name="kElByName" match="*" use="concat(generate-id(..), '+',name())"/> <xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="node()|@*"/> </xsl:copy> </xsl:template> <xsl:template match="*[starts-with(name(), 'ele')]"> <xsl:copy> <xsl:copy-of select="@*"/> <xsl:apply-templates select= "*[generate-id() = generate-id(key('kElByName', concat(generate-id(..), '+',name()) )[1]) ]" /> </xsl:copy> </xsl:template> </xsl:stylesheet> </pre> when applied on this XML document: <pre> <Root> <ele1> <child1>context1</child1> <child2>test1</child2> <child1>context1</child1> </ele1> <ele2> <child1>context2</child1> <child2>test2</child2> <child1>context2</child1> </ele2> <ele3> <child2>context2</child2> <child2>test2</child2> <child1>context1</child1> </ele3> </Root> </pre> produces the wanted result: <pre> <Root> <ele1> <child1>context1</child1> <child2>test1</child2> </ele1> <ele2> <child1>context2</child1> <child2>test2</child2> </ele2> <ele3> <child2>context2</child2> <child1>context1</child1> </ele3> </Root> </pre>
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1. POHow to remove duplicate XML nodes using XSLT
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1. USDimitre Novatchev
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