Move sub nodes into parent attributes with XSLT

Question

I have some XML which contains records and sub records, like this:

Answer 1

This might do it, run this following snippet of XSLT when processing the top level <record> elements:

<xsl:for-each select="record">
    <xsl:attribute name="{position()}">
        <xsl:value-of select="@string" />
    </xsl:attribute>
</xsl:for-each>

Essentially this iterates over each sub-<record> element and creates an <xsl:attribute> element describing the desired attribute. The position() function is called to get the relative position within the top level element: 1, 2, 3, etc.

This is not a complete solution; some familiarity with XSLT is assumed.

Answer 2

Here's a complete solution:

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

  <!-- By default, recursively copy all nodes unchanged -->
  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>

  <!-- But don't process any children of <record> (such as whitespace)... -->
  <xsl:template match="record/node()"/>

  <!-- ...except for doubly-nested records;
       convert them to attributes, named according to position -->
  <xsl:template match="record/record" priority="1">
    <xsl:variable name="pos">
      <xsl:number/>
    </xsl:variable>
    <xsl:attribute name="r{$pos}">
      <xsl:value-of select="@string"/>
    </xsl:attribute>
  </xsl:template>

</xsl:stylesheet>

Note that I changed the name of your attributes to "r1", "r2", etc., because XML doesn't allow you to start a name with a number.