Getting elements with default namespace (no namespace prefix) using XPath

问题

In this SOAP XML file, how can I get the 7 on a using a XPath query?

<soap:Envelope xmlns:soap="http://www.w3.org/2003/05/soap-envelope"
                            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                            xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <soap:Body>
        <HelloWorldResponse xmlns="http://tempuri.org/">
           <HelloWorldResult>7</HelloWorldResult>
        </HelloWorldResponse>
    </soap:Body>
</soap:Envelope>

This XPath query is not working //*[name () ='soap:Body'].

回答1:

If you have a namespace prefix set, you could use it, like:

//soap:Body

But since the nodes you are trying to get use a default namespace, without a prefix, using plain XPath, you can only acesss them by the local-name() and namespace-uri() attributes. Examples:

//*[local-name()="HelloWorldResult"]/text()

Or:

//*[local-name()="HelloWorldResult" and namespace-uri()='http://tempuri.org/']/text()

Or:

//*[local-name()="HelloWorldResponse" and namespace-uri()='http://tempuri.org/']/*[local-name()="HelloWorldResult"]/text()

To your xml, they will all give the same result, the text 7.

来源：https://stackoverflow.com/questions/16717211/getting-elements-with-default-namespace-no-namespace-prefix-using-xpath