Question

I have:

<nodes>
  <node>
    <name>node1</name>
    <other>stuff1</other>
    <node>
      <name>node11</name>
      <other>stuff11</other>
    </node>
    <node>
      <name>node12</name>
      <other>stuff12</other>
    </node>
  </node>
  <node>
    <name>node2</name>
    <other>stuff2</other>
  </node>
  <node>
    <name>node3</name>
    <other>stuff3</other>
  </node>
</nodes>

I want to end up with a flat structure like:

<nodes>
  <node>
    <name>node1</name>
    <other>stuff1</other>
  </node>
  <node>
    <name>node11</name>
    <other>stuff11</other>
  </node>
  <node>
    <name>node12</name>
    <other>stuff21</other>
  </node>
  <node>
    <name>node2</name>
    <other>stuff2</other>
  </node>
  <node>
    <name>node3</name>
    <other>stuff3</other>
  </node>
</nodes>

This is a simple example but I want to copy all elements in each node, but not the nested ‘node’ elements. I tried the copy-of to preserve the tags but that also preserves the nesting. I also tried copy but that leaves out all children.

Any ideas?

Answer 1

This should do what you’re after. For the top-level nodes element it applies templates for all descendant node elements in document order. For each node it copies all the non-node child elements.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

  <xsl:output method="xml" indent="yes" />

  <xsl:template match="node">
    <xsl:copy><xsl:copy-of select="*[local-name() != 'node']"/></xsl:copy>
  </xsl:template>

  <xsl:template match="/nodes">
    <xsl:copy><xsl:apply-templates select="descendant::node" /></xsl:copy>
  </xsl:template>
</xsl:stylesheet>