Retrieve value of attribute using XPath

Question

I am trying to retrieve the value of an attribute from an xmel file using XPath and I am not sure where I am going wrong..

This is the XML File

<soapenv:Envelope>
  <soapenv:Header>
    <common:TestInfo testID="PI1" />
  </soapenv:Header>
</soapenv:Envelope>

And this is the code I am using to get the value. Both of these return nothing..

XPathBuilder getTestID = new XPathBuilder("local-name(/*[local-name(.)='Envelope']/*[local-name(.)='Header']/*[local-name(.)='TestInfo'])");
XPathBuilder getTestID2 = new XPathBuilder("Envelope/Header/TestInfo/@testID");

Object doc2 = getTestID.evaluate(context, sourceXML);
Object doc3 = getTestID2.evaluate(context, sourceXML);

How can I retrieve the value of testID?

Which of the many `XPathBuilder` implementations are you using? — parsifal, Feb 08 '13 at 15:40
Please make sure to post *valid* XML, the ` closing tag actually was another opening one. — Jens Erat, Feb 08 '13 at 15:46
Possibly off-topic, but as this concerns SOAP: if you're trying to implement your own SOAP service/request parser, please don't; you'll cut yourself on one of its many many sharp edges. — akaIDIOT, Feb 08 '13 at 15:57

akaIDIOT · Answer 1 · 2013-02-08T15:55:46.793

0

The XML file you provided does not contain an <Envelope> element, so an expression that requires it will never match.

Post-edit edit

As can be seen from your XML snippet, the document uses a specific namespace for the elements you're trying to match. An XPath engine is namespace-aware, meaning you'll have to ask it exactly what you need. And, keep in mind that a namespace is defined by its uri, not by its abbreviation (so, /namespace:element doesn't do much unless you let the XPath engine know what the namespace namespace refers to).

edited Feb 08 '13 at 15:55

answered Feb 08 '13 at 15:20

akaIDIOT

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I was just showing a cross section as an example. Need help with the XPath expression.. – stinkyPete Feb 08 '13 at 15:27
1

While that is understandable, please always provide enough data for anyone here to reproduce the problem, should they decide to try. Asking people to help based on half a problem doesn't work well :) – akaIDIOT Feb 08 '13 at 15:52

score 0 · Answer 2 · answered Feb 08 '13 at 15:28

0

However you're iterating within the java, your context node is probably not what you think, so remove the "." specifier in your local-name(.) like so:

/*[local-name()='Header']/*[local-name()='TestInfo']/@testID worked fine for me with your XML, although as akaIDIOT says, there isn't an <Envelope> tag to be seen.

answered Feb 08 '13 at 15:28

JWiley

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I've added the Envelope tags to show that I do have them. The XPath string still isn't working with my code.. – stinkyPete Feb 08 '13 at 15:36
There's no difference between `local-name()` and `local-name(.)`. The first one is just one character shorter. – JLRishe Feb 08 '13 at 18:03

Jens Erat · Answer 3 · 2013-02-08T18:27:39.777

0

If you don't care for the namespaces and use an XPath 2.0 compatible engine, use * for it.

//*:Header/*:TestInfo/@testID

will return the desired input.

It will probably be more elegant to register the needed namespaces (not covered here, depends on your XPath engine) and query using these:

//soapenv:Header/common:TestInfo/@testID

edited Feb 08 '13 at 18:27

answered Feb 08 '13 at 15:43

Jens Erat

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That's not valid XPath. You can't use `*` as a namespace prefix wildcard. – JLRishe Feb 08 '13 at 18:04
You're right, it isn't valid XPath 1.0, but [allowed in XPath 2.0](http://developer.marklogic.com/blog/namespace-wildcards-in-xpath). I added this to my answer, thank you. – Jens Erat Feb 08 '13 at 18:26

score 0 · Answer 4 · edited May 23 '17 at 12:27

Your first XPath has an extra local-name() wrapped around the whole thing:

local-name(/*[local-name(.)='Envelope']/*[local-name(.)='Header']
                                                         /*[local-name(.)='TestInfo'])

The result of this XPath will either be the string value "TestInfo" if the TestInfo node is found, or a blank string if it is not.

If your XML is structured like you say it is, then this should work:

/*[local-name()='Envelope']/*[local-name()='Header']/*[local-name()='TestInfo']/@testID

But preferably, you should be working with namespaces properly instead of (ab)using local-name(). I have a post here that shows how to do this in Java.

Retrieve value of attribute using XPath

4 Answers4