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Home/ Questions/Q 494273
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T05:28:54+00:00 2026-05-13T05:28:54+00:00

It has been a great challenge for me to code XSLT to bring my

  • 0

It has been a great challenge for me to code XSLT to bring my desired output ..

Here is my XML input .. which I am testing on ..

<aaa id="1">
  <bbb id="2">text1</bbb>
  <ccc>text2</ccc>
  <ddd id="3">
    <eee att="4d">text3</eee>
    <fff>
      <ggg att="3d">
        <yyy>text4</yyy>
        <iii attr="jj"/>
      </ggg>
    </fff>
    <sss>
      <ttt info="Y">
        <nnn delete="N">
          <mmm>text5</mmm>
        </nnn>
      </ttt>
    </sss>
  </ddd>
</aaa>

The desired output :

<root>
  <aaa id="1"/>
  <bbb id="2">text1</bbb>
  <ccc>text2</ccc>
  <ddd id="3"/>
  <eee att="4d">text3</eee>
  <fff/>
  <ggg att="3d"/>
  <yyy>text4</yyy>
  <iii attr="jj"/>
  <sss/>
  <ttt info="Y"/>
  <nnn delete="N"/>
  <mmm>text5</mmm>
</root>

I am not able to write generalised code .. I mean, My code must be able to give similar output for different XML file with different tag names ..

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T05:28:54+00:00Added an answer on May 13, 2026 at 5:28 am

    Something like this?

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