How to parse XML using XmlPullParser

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;.

In the future, if you have a specific requirement (including but not limited to a particular parser), please mention it in the first draft of your question. — Pops
– Pops, Commented Nov 18, 2011 at 22:25

bdoughan · Accepted Answer · 2011-11-04 14:04:31Z

9

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);
    }

}

answered Nov 4, 2011 at 14:04

bdoughan

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Comments

ChadNC · Accepted Answer · 2011-11-04 14:23:23Z

5

Everyone has thier own preferences for XML processing in Java and my preference is to use JAXB when dealing with XML in Java because I find it easier to use that straight xpath.

answered Nov 4, 2011 at 14:23

ChadNC

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1 Comment

ChadNC Over a year ago

I would still use JAXB with the only difference being that I would use a transform to save the xml string as an xml file then unmarshall the file into a Java object using JAXB and then get what I wanted from the object.

hovanessyan · Accepted Answer · 2011-11-04 14:04:06Z

1

You can do that using Java Xpath API

answered Nov 4, 2011 at 14:04

hovanessyan

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Comments

shift66 · Accepted Answer · 2011-11-04 14:29:38Z

0

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".

answered Nov 4, 2011 at 14:29

shift66

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