CODEWITHSUNDEEP
xmlxslt
Subramanyam
Subramanyam

Reputation: 11

using xslt how to replace the node

<h:body>
<group id = "1">
<name>xxx</name>
<age>12</age>
<group id = "2">
<name>yyy</name>
<age>13</age>
</h:body>

using XSLT i want to replace id=1 with <group appearance = "field-list">

Upvotes: 0

Views: 64

Answers (1)

kjhughes
kjhughes

Reputation: 111726

Use the identity transform, overriding the part you wish to replace with something different. To wit:

Given this input XML document:

<h:body xmlns:h="http://example.org/h">
  <group id = "1">
    <name>xxx</name>
    <age>12</age>
  </group>
  <group id = "2">
    <name>yyy</name>
    <age>13</age>
  </group>
</h:body>

This XSLT transformation:

<xsl:stylesheet version="1.0"
                xmlns:h="http://example.org/h"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output omit-xml-declaration="yes" indent="yes"/>

  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="group[@id = '1']">
    <group appearance = "field-list">
      <xsl:apply-templates select="node()|@*"/>
    </group>
  </xsl:template>
</xsl:stylesheet>

Will produce this output XML document:

<h:body xmlns:h="http://example.org/h">
  <group appearance="field-list" id="1">
      <name>xxx</name>
      <age>12</age>
  </group>
  <group id="2">
      <name>yyy</name>
      <age>13</age>
  </group>
</h:body>

Upvotes: 1

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