XPathExpression.evaluate using Node

Asked Jun 05 '17 at 13:55

Active Jun 05 '17 at 13:55

Viewed 25 times

I have the following XML (which I receive and so do not have control of):

<root>
   <elementA  ....>
       <elementB  ....>
            <nsx:elementC  ...>

At a point in my code I have the Node - elementA. What I am trying to do is get elementC ignoring the namespaces (the document was parsed with namespace aware = true because I want to be able to use getLocalName throughout my code.

I am using the following code:

Element elementA = ...;
XPath xpath = XPathFactory.newInstance().newXPath();
XPathExpression expression = xpath.compile(<see below for xpath);
Node result = expression.evaluate(elementA, XPathConstants.NODE);

Here are my results per tried XPath:

"." --> elementA
"./*" --> elementB
"./elementB" --> null
"elementB" --> null

I don't get why if "./*" gives me elementB, "./elementB" gives me null.

Eventually, what I would like is something like: "./:elementB/:elementC"

asked Jun 05 '17 at 13:55

John B

32,493
6
77
98

I'm looking into my crystal ball here and am going to predict that your XML has a default namespace that you are ignoring. – Tomalak Jun 05 '17 at 13:58
have you tried ./*:elementB ? – Carlos Jun 05 '17 at 14:04
@Carlos "./*:elementB" results in an exception: "Prefix must resolve to a namespace" – John B Jun 05 '17 at 14:09
@Tomalak You are correct. Didn't notice. Is there an easy way to ignore? I have tried ```./*[local-name()=elementB]``` with no luck – John B Jun 05 '17 at 14:11
Don't ever ignore namespaces. Every XPath engine has a way to declare and use them. – Tomalak Jun 05 '17 at 14:15

XPathExpression.evaluate using Node

0 Answers0