CODEWITHSUNDEEP
xmlxslt
Unmesh Kondolikar
Unmesh Kondolikar

Reputation: 9312

How to output duplicate elements using XSLT?

I have xml which looks something like this -

<Root>
  <Fields>
    <Field name="abc" displayName="aaa" />
    <Field name="pqr" displayName="ppp" />
    <Field name="abc" displayName="aaa" />
    <Field name="xyz" displayName="zzz" />
  </Fields>
</Root>

I want the output to contain only those elements which have a repeating name-displayName combination, if there are any - 

<Root>
      <Fields>
        <Field name="abc" displayName="aaa" />
        <Field name="abc" displayName="aaa" />
      </Fields>
</Root>

How can I do this using XSLT?

Upvotes: 6

Views: 7070

Answers (2)

Dimitre Novatchev
Dimitre Novatchev

Reputation: 243449

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="kFieldByName" match="Field"
  use="concat(@name, '+', @displayName)"/>

 <xsl:template match=
  "Field[generate-id()
        =
         generate-id(key('kFieldByName',
                     concat(@name, '+', @displayName)
                     )[2])
        ]
  ">
     <xsl:copy-of select=
     "key('kFieldByName',concat(@name, '+', @displayName))"/>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document:

<Root>
    <Fields>
        <Field name="abc" displayName="aaa" />
        <Field name="pqr" displayName="ppp" />
        <Field name="abc" displayName="aaa" />
        <Field name="xyz" displayName="zzz" />
    </Fields>
</Root>

produces the wanted result:

<Field name="abc" displayName="aaa"/>
<Field name="abc" displayName="aaa"/>

Explanation:

  1. Muenchian grouping using composite key (on the name and displayName attributes).

  2. The only template in the code matches any Field element that is the second in its corresponding group. Then, inside the body of the template, the whole group is output.

  3. Muenchian grouping is the efficient way to do grouping in XSLT 1.0. Keys are used for efficiency.

  4. See also my answer to this question.

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:template match="/">
     <xsl:for-each-group select="/*/*/Field"
          group-by="concat(@name, '+', @displayName)">
       <xsl:sequence select="current-group()[current-group()[2]]"/>
   </xsl:for-each-group>
 </xsl:template>
</xsl:stylesheet>

When this transformation is applied on the provided XML document (shown above), again the wanted, correct result is produced:

<Field name="abc" displayName="aaa"/>
<Field name="abc" displayName="aaa"/>

Explanation:

  1. Use of <xsl:for-each-group>

  2. Use of the current-group() function.

Upvotes: 9

Jeff Yates
Jeff Yates

Reputation: 62377

To find duplicates, you need to iterate the Field elements and for each one, look for the set of Field elements in the whole document that have matching name and displayName attribute values. If the set has more than 1 element, you add that element into the output.

Here is an example of a template that achieves this:

<xsl:template match="Field">
    <xsl:variable name="fieldName" select="@name" />
    <xsl:variable name="fieldDisplayName" select="@displayName" />
    <xsl:if test="count(//Field[@name=$fieldName and @displayName=$fieldDisplayName]) > 1">
        <xsl:copy-of select="."/>
    </xsl:if>
</xsl:template>

Executing this template (wrapped in an appropriate XSLT file) on your sample data gives the following output:

<?xml version="1.0" encoding="utf-8"?>
<Root>
  <Fields>
    <Field name="abc" displayName="aaa" />
    <Field name="abc" displayName="aaa" />
  </Fields>
</Root>

Upvotes: 1

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