Thanks. I got it to work using w3c.dom with the following code (in case
anyone interested)...
private void processChildren(NodeList nList) {
for (int i=0; i<nList.getLength(); i++) {
// pass each node into traverse for processing
_attrib_name = traverse(nList.item(i));
String nodeStr = _attrib_name;
Node node = nList.item(i);
do {
node = node.getParentNode();
if (node!=null) {
nodeStr = node.getNodeName()+"."+nodeStr;
} // end if
} while(node!=null);
System.out.println("node path = " + nodeStr);
} // end for
}
----- Original Message -----
From: "Joost Diepenmaat" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Friday, April 08, 2005 1:01 PM
Subject: Re: [Developers] fully-qualified xml fields
> On Fri, Apr 08, 2005 at 12:26:55PM +0200, Emile wrote:
> > I'm referring to the actual XML document, in particular the class names.
> >
> > Given the following example:
> >
> > <grandad name="gramps">
> > <parents name="mom">
> > <kids name="John">
> > <attribute name="haircolour" value="blonde">
> > </kids>
> > </parents>
> > <parents name="dad">
> > <kids name="Jill">
> > <attribute name="haircolour" value="brunette">
> > </kids>
> > </parents>
> > </grandad>
> >
> > I'm trying to list the attrributes as:
> > gramps.mom.John.haircolour = "blonde"
> >
> > Regards
> > Emile
> >
>
> that depends on the kind of parser you're using.
>
> pseudo-code for DOM level 1:
>
> value = node.getAttribute("haircolour");
> name = "haircolour";
> do {
> node = node.parentNode;
> name = node.getAttribute("name")+"."+name;
> } while (node);
>
>
> J.
>
>
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>
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