Re: XPath with default namespace - No result

Olivier,

As Gareth mentioned in your JIRA post, what's happening is that the 
<report> and <account_number> elements are still bound to the 
"http://www.mydomain.com/report" namespace, even though they are not 
using a namespace prefix. Your XML is using a declared default namespace 
of "http://www.mydomain.com/report".

Therefore your XPath query will not return anything because it is asking 
for <report> and <account_number> elements that are in no namespace 
(because you haven't defined a namespace context), but your input XML 
only contains elements in the "http://www.mydomain.com/report" namespace.

If you can't control your input XML, then you need to set your namespace 
context so that the XPath query knows you are looking for <report> and 
<account_number> in the "http://www.mydomain.com/report" namespace. 
These pages should provide good instructions and hints:

http://xml.apache.org/xalan-j/xpath_apis.html#namespacecontext
http://www.edankert.com/defaultnamespaces.html

Good luck,
Nathan

Olivier Roger wrote:
> Hi, 
>   
> I am having difficulties to use XPath query on a XML dataset having its root
> element with an *xmlns attribute with no prefix*. 
>   
> In that specific case, the XPath query does not return the value. Since I
> have no prefix for the namespace I simply can get the value. 
> It this a known issue or am I simply doing something wrong ? 
>
> Here is my XML : 
>
> {noformat:title=default-ns.xml} 
> <?xml version="1.0" encoding="UTF-8"?> 
>
> <report xmlns="http://www.mydomain.com/report"> 
>     <account_number>123</account_number> 
> </report> 
> {noformat} 
>
> Here is my test method:
>
> {code} 
>     @Test 
>     public void xPathDefaultNS() throws XPathExpressionException { 
>         javax.xml.xpath.XPathFactory factory =
> javax.xml.xpath.XPathFactory.newInstance(); 
>         javax.xml.xpath.XPath xpath = factory.newXPath(); 
>         javax.xml.xpath.XPathExpression expression =
> xpath.compile("/report/account_number"); 
>         String result = expression.evaluate(new
> org.xml.sax.InputSource(this.getClass().getResourceAsStream("default-ns.xml"))); 
>         Assert.assertTrue("123".equals(result)); 
>     } 
> {code}
>
> From what Gareth responded on the Jira issue
> (https://issues.apache.org/jira/browse/XALANJ-2549) there seems to be a
> missing step (resolving the namespace). How I that done exactly ? 
>
>
>   
-- 
Nathan Nadeau
[email protected]
Software Development
Gleim Publications, Inc.
http://www.gleim.com