Arvind Singh
Arvind Singh

Reputation: 802

Need to fetch distinct elements through XSLT

I have a list of table elements under element named Query that may duplicate, I need to fetch distinct table elements (not its value but the tag/element name itself.

/ShopArea/Connection/Query/* lists the table names including duplicates.

Below is the XML

<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="ShopArea.xslt"?>
<ShopArea>
<Connection name="Connection1">
    <Report date="25-09-2011">

        <Query id="1">
            <TABLE1>1.1</TABLE1>
            <TABLE2>1.2</TABLE2>
            <TABLE3>1.3</TABLE3>
        </Query>
        <Query id="2">
            <TABLE21>2.1</TABLE21>
            <TABLE22>2.2</TABLE22>
            <TABLE23>2.3</TABLE23>
        </Query>
    </Report>

    <Report date="26-09-2011">
        <Query id="1">
            <TABLE1>26 1.1</TABLE1>
            <TABLE2>26 1.2</TABLE2>
            <TABLE3>26 1.3</TABLE3>
        </Query>
        <Query id="2">
            <TABLE21>26 2.1</TABLE21>
            <TABLE22>26 2.2</TABLE22>
            <TABLE23>26 2.3</TABLE23>
        </Query>
    </Report>
</Connection>
</ShopArea>

List of elements including duplicates

I refered How to select unique nodes in XSLT but I'm not able to get it right.

Upvotes: 1

Views: 290

Answers (1)

Dimitre Novatchev
Dimitre Novatchev

Reputation: 243599

I. XSLT 1.0. This transformation uses simple Muenchian grouping:

<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="kChildByName" match="Query/*"
  use="name()"/>

 <xsl:template match=
  "Query/*
       [generate-id()
       =
        generate-id(key('kChildByName', name())[1])
       ]">
     <xsl:value-of select="name()"/>
     <xsl:text>&#xA;</xsl:text>
 </xsl:template>

 <xsl:template match="text()"/>
</xsl:stylesheet>

When applied on the provided XML document:

<ShopArea>
    <Connection name="Connection1">
        <Report date="25-09-2011">
            <Query id="1">
                <TABLE1>1.1</TABLE1>
                <TABLE2>1.2</TABLE2>
                <TABLE3>1.3</TABLE3>
            </Query>
            <Query id="2">
                <TABLE21>2.1</TABLE21>
                <TABLE22>2.2</TABLE22>
                <TABLE23>2.3</TABLE23>
            </Query>
        </Report>
        <Report date="26-09-2011">
            <Query id="1">
                <TABLE1>26 1.1</TABLE1>
                <TABLE2>26 1.2</TABLE2>
                <TABLE3>26 1.3</TABLE3>
            </Query>
            <Query id="2">
                <TABLE21>26 2.1</TABLE21>
                <TABLE22>26 2.2</TABLE22>
                <TABLE23>26 2.3</TABLE23>
            </Query>
        </Report>
    </Connection>
</ShopArea>

the wanted, correct result (all different names of elements that are children of a Query) is produced:

TABLE1
TABLE2
TABLE3
TABLE21
TABLE22
TABLE23

II. XSLT 2.0: This transformation uses <xsl:for-each-group>:

<xsl:stylesheet version="2.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema">
    <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/*/*">
     <xsl:for-each-group select="Report/Query/*"
                         group-by="name()">
      <xsl:sequence select="current-grouping-key(), '&#xA;'"/>
     </xsl:for-each-group>
 </xsl:template>

 <xsl:template match="text()"/>
</xsl:stylesheet>

Upvotes: 1

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