Hi ,
I tried the sample code that you provided and it seems to be working. I
added a couple of snippets to print out the nodes that were being returned:
After the query:
node = XPathAPI.selectSingleNode(xmlDoc, "//[EMAIL PROTECTED]'10099']/*");
I tried:
System.out.println("Node: " + node.getNodeName());
and was returned:
Node: DemoInfo
After the second query:
NodeIterator nl = XPathAPI.selectNodeIterator(node, "./*");
I tried:
Node n = null;
while ((n = nl.nextNode()) != null) {
System.out.println("Got node: " + n.getNodeName());
}
and was returned:
Got node: Age
Got node: Gender
Is this different from the results you were seeing or expecting?
Ilene.
"Pae Choi"
<[EMAIL PROTECTED] To: <[EMAIL PROTECTED]>,
Joseph Kesselman/Watson/[EMAIL PROTECTED]
k.net> cc:
Subject: Re: 1st parameter,
"Node context Node" in the XPathAPI's methods
11/25/2002 01:00
PM
I was hoping that, but did not work in that way. I have posted
source code, including the XML doc. If you see the Java code and
better to test it to see the result.
Pae
> From the Javadoc:
> contextNode - The node to start searching from.
>
> In other words, the note where evaluation of the XPath should start.
>
> ______________________________________
> Joe Kesselman / IBM Research
