Is it possible to find the position of a node outside of the current context? For instance:

<ROOT>
  <PARENTTYPE1>
    <CHILD id="id1"/>
    <CHILD id="id2"/>
  </PARENTTYPE1>
  <PARENTTYPE2>
    <CHILD id="id3"/>
    <CHILD id="id4"/>
  </PARENTTYPE2>
</ROOT>

<xsl:template name="findPositionOfChild">
  <xsl:param name="idOfChild"/>
  <xsl:for-each select="//CHILD">
    <xsl:if test="@id=$idOfChild">
      <xsl:value-of select="position()"/>
    </xsl:if>
  </xsl:for-each>
</xsl:template>

If I call the above template with <xsl:with-param name="idOfChild">id3</xsl:with-param> the result of the position() call will be "3". I assume that this is because the position of the child with id="id3" in the node set returned by <xsl:for-each select="//CHILD"> is 3. But I want to find the position of the child with id="id3" in the context of that child's parent (it's the 1st child of the parent <PARENTTYPE2>).

I assume this can be done with an xpath expression similar to the following:
<xsl:variable name="id3Var">id3</xsl:variable>
<xsl:variable name="id3Pos" select="//*[@id=$id3Var].position()"/>
But the syntax is incorrect.

The position() function returns the position of a node within the sequence of nodes currently being processed. This has nothing to do with its position in the source tree.

To find out the position of a node in the source tree, use xsl:number or count(preceding-sibling::*), or suitable variations.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Ah, I was under the impression that preceding-sibling::* worked the same way as position() did. i.e that it applies to the nodes being processed instead of the source tree.

Thanks