如何使用XmlPullParser解析XML

Question

I want to use the Java XmlPullParser to parse an XML file like this:

<start>
    <tag1> text1 </tag1>
    <tag2> 
        <tag3>text3</tag3>
        <tag1>text4</tag1>
    </tag2>
    <tag4> text5</tag4>
</start>

I want only <tag1> text1 </tag1> as the result. My current method produces both <tag1> text1 </tag1> and <tag1>text4<tag1/> . What else must I do?

Edit:

I am getting the XML in String format. I want to parse it using org.xmlpull.v1.XmlPullParser; .

Answer 1

You could also just use the javax.xml.xpath APIs:

import java.io.FileReader;
import javax.xml.xpath.*;
import org.w3c.dom.Node;
import org.xml.sax.InputSource;

public class XPathDemo {

    public static void main(String[] args) throws Exception {
        XPathFactory xpf = XPathFactory.newInstance();
        XPath xpath = xpf.newXPath();
        InputSource xml = new InputSource(new FileReader("input.xml"));
        Node result = (Node) xpath.evaluate("/tag1", xml, XPathConstants.NODE);
        System.out.println(result);
    }

}

Answer 2

每个人都有自己对Java中XML处理的偏好，我更喜欢在Java中处理XML时使用JAXB ，因为我发现使用这个直接的xpath更容易。

Answer 3

您可以使用Java Xpath API执行此操作

Answer 4

try to use dom4j library

    InputStream is = FileUtils.class.getResourceAsStream(filepath);
    SAXReader reader = new SAXReader();
    org.dom4j.Document doc = reader.read(is);
    is.close();
    Element content = doc.getRootElement();
    List<Element> els = content.elements("elemeNtname");

after this in method you will get in els list all the elements in xml file with name "elemeNtname".