How to move XML element at end

krish · December 15th, 2009, 01:43 AM

Hi,

I am having XML elements like below:

<1>
<2>text 1</2>
<3>text 2</3>
<2>text 3</2>
<2>text 4</2>
<3>text 5</3></1>

I want to move the XML element <3> at end like below:

<1>
<2>text 1</2>
<2>text 3</2>
<2>text 4</2>
</1><3>text 2</3>
<3>text 5</3>

Can any one help me on how to achive this.

Thanks,
Krish

samjudson · December 15th, 2009, 05:13 AM

The example output above is not a valid XML document as it has no single root element. Is the <3> element meant to be moved outside the <1> element?

mhkay · December 15th, 2009, 06:33 AM

It helps to use well-formed XML examples: you can't use numbers as XML element names, which means I can't show you how to write path expressions to manipulate such elements, let alone testing that the examples work.

But basically you want a template rule like this:

Code:

<xsl:template match="one">
  <one>
     <xsl:copy-of select="two"/>
  </one>
  <xsl:copy-of select="three"/>
</xsl:template>

mrame · December 15th, 2009, 06:58 AM

With XSLT 2.0 and the input xml as:

Code:

<root>
<test2>text 1</test2>
<test3>text 2</test3>
<test2>text 3</test2>
<test2>text 4</test2>
<test3>text 5</test3>
</root>

<p>I have given below two snippets. First will put <test3> as the last child in <root> and the next snippet will put <test3> out of <root></p>

Code:

<xsl:template match="root">
      <xsl:copy>
   <xsl:copy-of select="* except test3"/>
   <xsl:copy-of select="test3"/>
   </xsl:copy>
  </xsl:template>

Code:

  <xsl:template match="root">
      <xsl:copy>
   <xsl:copy-of select="* except test3"/>
   </xsl:copy>
   <xsl:copy-of select="test3"/>
  </xsl:template>