You could try something like this. Note: the context of the sorting
XPath is the XPath's result.
try {
SAXReader reader = new SAXReader();
Document document = reader.read( "file:sort.xml");
XPath xpath = DocumentHelper.createXPath("//item");
XPath sortXPath = DocumentHelper.createXPath("@name");
List nodes = xpath.selectNodes(document, sortXPath);
for ( int i = 0; i < nodes.size(); i++) {
System.out.println( ((Node)nodes.get(i)).asXML());
}
} catch ( DocumentException e) {
// the document is not well-formed.
e.printStackTrace();
}
The XML used was the following:
<items>
<item name="bbb"/>
<item name="ccc"/>
<item name="aaa"/>
</items>
Regards,
Edwin
--
http://www.edankert.com/
-------------------------------------------------------------------------
This SF.net email is sponsored by DB2 Express
Download DB2 Express C - the FREE version of DB2 express and take
control of your XML. No limits. Just data. Click to get it now.
http://sourceforge.net/powerbar/db2/
_______________________________________________
dom4j-user mailing list
[email protected]
https://lists.sourceforge.net/lists/listinfo/dom4j-user
