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copied!<p>First off: You can't use XPath to get an XML document "down to something". You can use it to <em>select</em> nodes, that's all. If you want to change the XML document, use XSLT.</p> <p>This expression:</p> <pre><code>//item/name/parent::item </code></pre> <p>does not select "the name nodes, and its parent", it selects the parent nodes of <code><name></code> nodes, and nothing else. </p> <p>Strictly speaking, it selects all <code><item></code> nodes that happen to be parent of a <code><name></code> node that is itself child of an <code><item></code> node. Which is equivalent to using just <code>"//item"</code>, when you think about it.</p> <p>There is no way to select a structure of nodes. You can only select a list of nodes - a node set. You could then traverse those nodes and find out about their position in the document, but the node set itself is flat.</p> <p>I think you need to explain more closely what you are trying to do. I could write an XSL transformation that does what you seem to intend, but unless I am sure what you intend... ;-)</p> <p>EDIT:</p> <p>Here is one minimalistic XSLT 1.0 approach that would do it.</p> <pre><code><xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="root | item | name | timestamp"> <xsl:copy> <xsl:apply-templates select="*" /> <xsl:if test="count(*) = 0"> <xsl:value-of select="text()" /> </xsl:if> </xsl:copy> </xsl:template> <xsl:template match="* | text()" /> </xsl:stylesheet> </code></pre> <p>Output for your sample (indentation mine):</p> <pre><code><root>e <item> <name>item name</name> <timestamp>47398743598</timestamp> </item> <item> ... </item> </root> </code></pre>

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