XPath Expression for getting default namespace

Question

I am looking for an XPath Expression to get the default namespace of a node.

My XML document might look like this.

<aaa>
    <bbb xmlns="ns1" xmlns:ggg="ns2" attribute="value" attribute2="value2">
        <ccc />
    </bbb>
    <bbb xmlns="ns4" xmlns:ggg="ns5" attribute="value" attribute2="value2">
        <ddd />
    </bbb>
</aaa>

I want to get ns1. I've tried it with the following expressions but none of it worked (I'm programming in Java 1.7 (so it is XPath 1.0) and tested with the online XPath visualizer on http://chris.photobooks.com/xml/default.htm).

/aaa/*[name()='bbb' and position()=1]/attribute::*

This expression only selects the attribute- and the attribute2-nodes.

/aaa/*[name()='bbb' and position()=1]/namespace::*

This expression does not select anything.

I would be grateful if someone has a hint for me.

Michael Kay · Accepted Answer · 2013-12-10 00:02:02Z

4

Your expression

/aaa/*[name()='bbb' and position()=1]/namespace::*

is correct and returns three namespace nodes. The problem might be in the way you are processing these nodes after they are returned. The expression should work in both XPath 1.0 and XPath 2.0, though I haven't checked it with the XPath engine built in to the JDK.

(Incidentally the notion that because you are using JDK 1.7 therefore you are using XPath 1.0 is a complete non sequitur, since there are several XPath 2.0 engines available for Java users).

To return only the namespace URI corresponding to the default namespace, use

/aaa/*[name()='bbb' and position()=1]/namespace::*[name()='']

Or indeed, since this query already assumes that the bbb element is in the default namespace, use

namespace-uri(/aaa/*[name()='bbb' and position()=1])

answered Dec 10, 2013 at 0:02

Michael Kay

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1 Comment

Anna Karina-Fan Over a year ago

Thanks for the answer. I'm glad I did not make a mistake. It really was the build in XPath engine. Now I'm using saxon library and everything works fine.

jgridley · Accepted Answer · 2013-12-09 20:13:55Z

1

This page explains the namespace-uri function in xpath.

It looks like you want something like this:

namespace-uri(/aaa/*[name()='bbb' and position()=1])

If the namespace isn't specified, you will get an empty string as the result.

EDIT: I reread your question and am not sure my answer is what you are looking for. The namespace-uri function is used to find the uri for nodes like <ns1:bbb>. It would return the uri for ns1. I'm not sure there is a way to use x-path to look at namespace definitions other than when they are actually used.

edited Dec 9, 2013 at 20:13

answered Dec 9, 2013 at 20:01

jgridley

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1 Comment

Anna Karina-Fan Over a year ago

Hi, you did understand me right. That is a good solution, although like my query it does not work unter build in XPath engine. A change to saxon library worked fine. Thanks!

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