java - java 中命名空间 xpath 的输出

Question

我有以下代码，但在特定字段及其输出方面遇到了一些问题。命名空间已连接，但似乎未在必填字段上输出。任何有关这方面的信息都会很棒。

import org.w3c.dom.Document;
import org.xml.sax.SAXException;

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPathFactory;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathExpressionException;
import java.io.ByteArrayInputStream;
import java.io.IOException;

public class test
{
 public static void main(String args[])

 {


     String xmlStr =    "<aws:UrlInfoResponse xmlns:aws=\"http://alexa.amazonaws.com/doc/2005-10-05/\">\n" +
                "    <aws:Response xmlns:aws=\"http://awis.amazonaws.com/doc/2005-07-11\">\n" +
                "        <aws:OperationRequest>\n" +
                "            <aws:RequestId>blah</aws:RequestId>\n" +
                "        </aws:OperationRequest>\n" +
                "        <aws:UrlInfoResult>\n" +
                "            <aws:Alexa>\n" +
                "                <aws:TrafficData>\n" +
                "                    <aws:DataUrl type=\"canonical\">harvard.edu/</aws:DataUrl>\n" +
                "                    <aws:Rank>1635</aws:Rank>\n" +
                "                </aws:TrafficData>\n" +
                "            </aws:Alexa>\n" +
                "        </aws:UrlInfoResult>\n" +
                "        <aws:ResponseStatus xmlns:aws=\"http://alexa.amazonaws.com/doc/2005-10-05/\">\n" +
                "            <aws:StatusCode>Success</aws:StatusCode>\n" +
                "        </aws:ResponseStatus>\n" +
                "    </aws:Response>\n" +
                "</aws:UrlInfoResponse>";

                    DocumentBuilderFactory xmlFact = DocumentBuilderFactory.newInstance();

                    xmlFact.setNamespaceAware(true);

        DocumentBuilder builder = null;
        try {
            builder = xmlFact.newDocumentBuilder();
        } catch (ParserConfigurationException e) {
            e.printStackTrace();       }

        Document doc = null;
        try {
            doc = builder.parse(

     new ByteArrayInputStream( xmlStr.getBytes()));
        } catch (SAXException e) {
            e.printStackTrace();     } catch (IOException e) {
            e.printStackTrace();       }






     System.out.println(doc.getDocumentElement().getNamespaceURI());
     System.out.println(xmlFact.isNamespaceAware());

  String xpathStr = "//aws:OperationRequest";

            XPathFactory xpathFact = XPathFactory.newInstance();

            XPath xpath = xpathFact.newXPath();

        String result = null;
        try {
            result = xpath.evaluate(xpathStr, doc);
        } catch (XPathExpressionException e) {
            e.printStackTrace();  //To change body of catch statement use File | Settings | File Templates.
        }

        System.out.println("XPath result is \"" +  result + "\"");


}
}

Answer 1

Xpath 中的命名空间匹配不仅仅是匹配前缀的字符串。您必须实际定义一个 NamespaceContext 对象并将其与 xPath 关联。如果文档中的前缀与 xPath 中的前缀相同，实际上根本不重要

private NamespaceContext ns = new NamespaceContext() {
public String getNamespaceURI(String prefix) {
if (prefix.equals("ns1") return "http://alexa.amazonaws.com/doc/2005-10-05/";
else return XMLConstants.NULL_NS_URI;
}
public String getPrefix(String namespace) {
throw new UnsupportedOperationException();
}
public Iterator getPrefixes(String namespace) {
throw new UnsupportedOperationException();
}};

XPathFactory xpfactory = XPathFactory.newInstance();
XPath xpath = xpfactory.newXPath();
xpath.setNamespaceContext(ns);

String xpathStr = "//ns1:OperationRequest";
//and so on