Hi, there.
I also needed such functionality badly, and ended up writing my own method.
As I would appreciate someone having a look at this and maybe distribution
with the Xalan XPath API.
Anyway, here is my code:
----------------------------------------8<-----------
/**
* Returns the position of the node within the list of the parent's
children.
* @param node The node whose position shall be calculated.
* @return Position of the node. The first node has position=1 so these
values can be used
* directly with XPath expressions.
*/
public static int getNodePosition(Node node) {
Node parent = node.getParentNode();
if(parent==null)
return 1;
NodeList nl = parent.getChildNodes();
for(int i=0;i<nl.getLength();i++) {
if(nl.item(i)==node)
return i+1;
}
return -1;
}
/**
* Returns the position of the node within the list of the parent's
children.
* This method counts only the nodes of the specified type.
* @param node The node whose position shall be calculated.
* @return Position of the node. The first node has position=1 so these
values can be used
* directly with XPath expressions.
*/
public static int getNodePosition(Node node, short nodetype) {
Node parent = node.getParentNode();
if(parent==null)
return 1;
NodeList nl = parent.getChildNodes();
int pos = nl.getLength()-1;
while(nl.item(pos)!=node)
pos--;
int count=pos-1;
while(count>=0) {
if(nl.item(count).getNodeType()!=nodetype)
pos--;
count --;
}
return pos+1; //index is 1-based
}
/**
* Returns the position of the node within the list of the parent's
children.
* This method counts only the nodes with the given node name.
* @param node The node whose position shall be calculated.
* @return Position of the node. The first node has position=1 so these
values can be used
* directly with XPath expressions.
*/
public static int getNodePosition(Node node, String nodeName) {
Node parent = node.getParentNode();
if(parent==null)
return 1;
NodeList nl = parent.getChildNodes();
int pos = nl.getLength()-1;
while(nl.item(pos)!=node)
pos--;
int count=pos-1;
while(count>=0) {
if(!nl.item(count).getNodeName().equals(nodeName))
pos--;
count --;
}
return pos+1; //index is 1-based
}
public static String getLocationPath(Node node) {
return getLocationPath(node, null);
}
public static String getLocationPath(Node node, Node context) {
if(node==null)
return null;
String result;
if(node==context)
return ".";
switch(node.getNodeType()) {
case Node.DOCUMENT_NODE:
result = "";
break;
case Node.ELEMENT_NODE:
result = getLocationPath(node.getParentNode(), context) +
"/"+node.getNodeName()+"["+getNodePosition(node, node.getNodeName())+"]";
break;
case Node.TEXT_NODE:
result = getLocationPath(node.getParentNode(), context) +
"/text()["+getNodePosition(node, org.w3c.dom.Node.TEXT_NODE)+"]";
break;
default:
result = getLocationPath(node.getParentNode(), context) +
"/*["+getNodePosition(node)+"]";
}
return result;
}
---------------------------------------->8-----------
> -----Urspr�ngliche Nachricht-----
> Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
> Gesendet: Dienstag, 27. November 2001 01:47
> An: [EMAIL PROTECTED]
> Betreff: Re: XPath for a node
>
>
>
> > Is there any support in Xerces API to get the XPath
> value for a node
> > in a XML document?
>
> There's been lots of discussion about providing an XPath
> generating utility
> function, but I don't think anyone has ever gotten around to
> contributing
> one. The basic logic I'd try would be something like the following
> (untested!) pseudocode:
>
> buffer=""
> while(thisnode.getNodeType()!=DOCUMENT_NODE)
> {
> count=0;
> name=thisnode.getNodeName();
> for(prevsib=thisnode.getPreviousSibling();
> prevsib!=null;
> prevsib=prevsib.getPreviousSibling())
> {
> if(name.equals(prevsib.getNodeName())
> ++count;
> }
> buffer="/"+name+"["+count+"]"+buffer;
> }
>
> As sketched that only works for elements; it'd have to be
> refined to deal
> with starting from other kinds of nodes. This sketch isn't
> namespace-aware;
> that'd also have to be fixed. And of course it assumes the
> node is in the
> main document tree; if it's part of an orphaned subtree or a
> DocumentFragment tree, you'd have to decide how best to
> represent that.
>
> A few days's hacking ought to be all that's needed, though.
> Good project
> for someone who's reasonably familiar with the DOM and wants
> to become more
> familiar with XPath to tackle and contribute to Xalan...
>
>
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