xpath to get the entire xml tree structure

Question

Is there a way using xpath to obtain the entire tree structure. For example, say this is the xml

<a>
   <b>
      <c>
         data
      </c>
   </b>
</a>

I would like the xpath to pick up all the contents of the node and the result should be

  <b>
     <c>
        data
     </c>
  </b>

I have been using VTD-XML and java so far to retrieve the elements so far. This is the code I have been using.

    VTDGen vg = new VTDGen();
    vg.setDoc(xmlString);
    vg.parse(true);
    VTDNav vn = vg.getNav();
    AutoPilot ap = new AutoPilot(vn);
    ap.selectXPath(xPath);
    String data = ap.evalXPathToString();

What exactly are you expecting it to do? Can you show some code? — John Saunders, Jan 13 '12 at 00:29
I am using VTD XML and Java to get xpaths. I have been able to get the normal xpaths using AutoPilot ap = new AutoPilot(); ap.selectXPath() ; String element = ap.evalXPathToString() ; — Ramesh Patil, Jan 13 '12 at 00:32
Take the time to read [ask]. Can you show the code you are using, and show what the result is, and show why the result is not what you expect. — John Saunders, Jan 13 '12 at 00:35
Sorry about that. This is my first post on stackoverflow. I have edited the main post with the code. — Ramesh Patil, Jan 13 '12 at 00:43
I think you should call getElementFragment(), if you tag this question with vtd-xml i could have answered it a lot earlier — vtd-xml-author, Aug 07 '13 at 22:57

score 4 · Answer 1 · answered Jan 13 '12 at 07:47

Remember that XPath selects a node or set of nodes. Think of it as returning pointers into your original source tree. It doesn't return "the entire tree structure", it returns pointers to selected nodes in that structure. Whatever you select, the whole structure is available to your program by navigating from the selected nodes to others. That navigation can be done using further XPath expressions, or using a DOM-like navigational API, or implicitly by operations such as serialization (typically when you serialize a node, it shows the whole subtree rooted at that node).

score 3 · Answer 2 · answered Jan 13 '12 at 04:03

Use:

/*/node()

This selects all children nodes of the top element of the XML document. The set of all these nodes has exactly the wanted subtree structure because every selected element preserves its entire sub-tree (of which this element is the top-node).

XSLT-based verification:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="/">
  <xsl:copy-of select="/*/node()"/>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the provided XML document:

<a>
  <b>
     <c>
      data
     </c>
  </b>
</a>

the XPath expression is evaluated and the selected nodes are output, producing the wanted, correct result:

<b>
   <c>
      data
     </c>
</b>

score 2 · Answer 3 · answered Jan 13 '12 at 00:46

2

Ideally, "//b" selects nodes in the document from the current node that match the selection.

answered Jan 13 '12 at 00:46

Sunil

1,238
3
13
21

I found this link that worked. I found this link which helped. [http://stackoverflow.com/a/6012662/1146690][1] [1]: http://stackoverflow.com/a/6012662/1146690 – Ramesh Patil Jan 13 '12 at 01:08

score 1 · Answer 4 · answered Nov 29 '13 at 06:58

1

Use

/descendant-or-self::node()

answered Nov 29 '13 at 06:58

Ravindranath Akila

209
3
35
45

xpath to get the entire xml tree structure

4 Answers4

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