XML: writing xslt to transform xmlelement into xmlattribute

There are several ways to do what you are asking. If you want a static method, then the following will work:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >

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

  <xsl:template match="/">
      <xsl:apply-templates />
  </xsl:template>

<xsl:template match="ROOT">
            <xsl:apply-templates />
      </xsl:template>


<xsl:template match="Element1">
      <Element1>
            <xsl:attribute name="Element2"><xsl:value-of select="Element2"/></xsl:attribute>
      </Element1>
</xsl:template>

</xsl:stylesheet>

You can actually match many elements and use the value of name() or local-name(), and grab the following element's value as an attribute value. Without knowing more about what your specific needs are, the above is a rudimentary but working example.

HTH,

kmartin7

Thank you kmartin7.
Existing XML is actually like this:
<ROOT>
   <Element1>
      <Element2>
         Value
      </Element2>
   </Element1>
   <Element3>Value</Element3>
   ...
   <ElementN>Value</ElementN>
</ROOT>

Element2 will be always the same. All other elements can have different names in different XML files. I need solution to be as generic as possible.

so you would like it to iterate to all nodes ?

quasar_ee:

Geert's solution will work for you, regarding your last comment.

kmartin7