如何在具有默认命名空间的xml文档上使用XPath [英] How to use XPath on xml docs having default namespace

问题描述

我想操作具有默认命名空间但没有前缀的xml doc。有没有办法使用没有命名空间uri的xpath，就好像没有命名空间一样？

我相信如果我们将documentBuilderFactory的namespaceAware属性设置为false，那么应该可行。但在我的情况下，它不起作用。

我的理解是不正确的还是我在代码中犯了一些错误？

I want to manipulate xml doc having default namespace but no prefix. Is there a way to use xpath without namespace uri just as if there is no namespace?
I believe it should be possible if we set namespaceAware property of documentBuilderFactory to false. But in my case it is not working.
Is my understanding is incorrect or I am doing some mistake in code?

这是我的代码：

    DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
    domFactory.setNamespaceAware(false);
    try {
        DocumentBuilder builder = domFactory.newDocumentBuilder();
        Document dDoc = builder.parse("E:/test.xml");

        XPath xPath = XPathFactory.newInstance().newXPath();
        NodeList nl = (NodeList) xPath.evaluate("//author", dDoc, XPathConstants.NODESET);
        System.out.println(nl.getLength());
    } catch (Exception e) {
        e.printStackTrace();
    }

这是我的xml：

<?xml version="1.0" encoding="UTF-8"?>
<root xmlns="http://www.mydomain.com/schema">
  <author>
    <book title="t1"/>
    <book title="t2"/>
  </author>
</root>

推荐答案

使用默认命名空间的文档的XPath处理（无前缀）与使用前缀的文档的XPath处理相同：

The XPath processing for a document that uses the default namespace (no prefix) is the same as the XPath processing for a document that uses prefixes:

对于命名空间限定文档，您可以在执行XPath时使用NamespaceContext。您需要在XPath中为片段添加前缀以匹配NamespaceContext。您使用的前缀不需要与文档中使用的前缀匹配。

For namespace qualified documents you can use a NamespaceContext when you execute the XPath. You will need to prefix the fragments in the XPath to match the NamespaceContext. The prefixes you use do not need to match the prefixes used in the document.

• http://download.oracle.com/javase/6/docs/api/javax/xml/ namespace / NamespaceContext.html

• http://download.oracle.com/javase/6/docs/api/javax/xml/namespace/NamespaceContext.html

以下是代码的外观：

import java.util.Iterator;
import javax.xml.namespace.NamespaceContext;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;

public class Demo {

    public static void main(String[] args) {
        DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
        domFactory.setNamespaceAware(true);
        try {
            DocumentBuilder builder = domFactory.newDocumentBuilder();
            Document dDoc = builder.parse("E:/test.xml");

            XPath xPath = XPathFactory.newInstance().newXPath();
            xPath.setNamespaceContext(new MyNamespaceContext());
            NodeList nl = (NodeList) xPath.evaluate("/ns:root/ns:author", dDoc, XPathConstants.NODESET);
            System.out.println(nl.getLength());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    private static class MyNamespaceContext implements NamespaceContext {

        public String getNamespaceURI(String prefix) {
            if("ns".equals(prefix)) {
                return "http://www.mydomain.com/schema";
            }
            return null;
        }

        public String getPrefix(String namespaceURI) {
            return null;
        }

        public Iterator getPrefixes(String namespaceURI) {
            return null;
        }

    }

}

注意：
我还使用了丹尼斯。

以下似乎也有效，并且更接近原来的问题：

The following also appears to work, and is closer to your original question:

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;

public class Demo {

    public static void main(String[] args) {
        DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
        try {
            DocumentBuilder builder = domFactory.newDocumentBuilder();
            Document dDoc = builder.parse("E:/test.xml");

            XPath xPath = XPathFactory.newInstance().newXPath();
            NodeList nl = (NodeList) xPath.evaluate("/root/author", dDoc, XPathConstants.NODESET);
            System.out.println(nl.getLength());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

}

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