使用Java中的XPath解析XML - 使用Java中的Xpath和NodeList从XML文件中获取数据

Ali*_*oud 5 java xml xpath nodelist

我有这个xml文件,我想用Xpath获取一些值.

工作的一半已经完成但我在文件的最后部分遇到了一些麻烦(States Node)

<?xml version="1.0" encoding="UTF-8"?>
<favoris>
    <workflow codewf="wf1000">
        <information>
            <title>wf1</title>
            <desc>description 1</desc>
            <nberState>2</nberState>
            <text>text text text text text text text</text>
        </information>
        <states>
            <state id="1" IDemployee="2">description1</state>
            <state id="2" IDemployee="3">description2</state>
        </states>
    </workflow>

    <workflow codewf="wf2000">
        <information>
            <title>wf2</title>
            <desc>description 2</desc>
            <nberState>3</nberState>
            <text>text text text text text text text</text>
        </information>
        <states>
            <state id="1" IDemployee="3">description1</state>
            <state id="2" IDemployee="2">description2</state>
            <state id="3" IDemployee="4">description2</state>
        </states>
    </workflow>

</favoris>
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这里是java代码:package myxml;

import java.io.FileReader;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;
import org.w3c.dom.Element;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
public class xmlParty {
  public static void main(String[] args) throws Exception {
    XPathFactory factory = XPathFactory.newInstance();
    XPath xPath = factory.newXPath();
    NodeList favoris = (NodeList) xPath.evaluate("/favoris/workflow[@codewf='wf1000']", 
            new InputSource(new FileReader("a.xml")), 
            XPathConstants.NODESET);
    for (int i = 0; i < favoris.getLength(); i++) {
      Element workflow = (Element) favoris.item(i);
      String title = xPath.evaluate("information/title", workflow);
      String desc_w = xPath.evaluate("information/desc", workflow);
      String nberState = xPath.evaluate("information/nberState", workflow);
      String text = xPath.evaluate("information/text", workflow);
      System.out.println(workflow.getAttribute("codewf") +" "+title + " " + desc_w + " " + nberState + " " + text );

      NodeList States = (NodeList)xPath.evaluate("states/state", workflow, XPathConstants.NODESET);
      System.out.println(States.getLength());
      for (int k = 0; k < States.getLength(); k++) {
          String desc_state = xPath.evaluate("states/state", workflow);
          System.out.println(desc_state ); 
      }


   }
  }
}
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输出将是:

第一个例子

wf1000 wf1 description 1 2 text text text text text text text
2
description1
description1
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第二个例子

wf2000 wf2 description 2 3 text text text text text text text
3
description1
description1
description1
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查看ID为2的状态,文本description2不是description1.我认为解析器不会移动到第二个孩子,它总是仍然在最小的孩子.那么我该怎么做以及如何做才能获得状态的属性????????

nab*_*tar 3

你将不得不做类似的事情:

for (int k = 0; k < States.getLength(); k++) {
          String desc_state = xPath.evaluate("states/state[position()=" + (k + 1) + "]", workflow);
          String id_employee = xPath.evaluate("states/state[position()=" + (k + 1) + "]/@IDemployee", workflow);
          System.out.println(desc_state + ":" + id_employee); 
}
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