parse an xml string in java?

Question

how do you parse xml stored in a java string object?

Java's XMLReader only parses XML documents from a URI or inputstream. is it not possible to parse from a String containing an xml data?

Right now I have the following:

try {
    SAXParserFactory factory = SAXParserFactory.newInstance();
    SAXParser sp = factory.newSAXParser();
    XMLReader xr = sp.getXMLReader(); 

    ContactListXmlHandler handler = new ContactListXmlHandler();
    xr.setContentHandler(handler);
    xr.p
} catch (ParserConfigurationException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} catch (SAXException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

And on my handler i have this:

public class ContactListXmlHandler extends DefaultHandler implements Resources {

    private List<ContactName> contactNameList = new ArrayList<ContactName>();

    private ContactName contactItem;

    private StringBuffer sb;

    public List<ContactName> getContactNameList() {
        return contactNameList;
    }

    @Override
    public void startDocument() throws SAXException {
        // TODO Auto-generated method stub
        super.startDocument();

        sb = new StringBuffer();
    }

    @Override
    public void startElement(String uri, String localName, String qName,
            Attributes attributes) throws SAXException {
        // TODO Auto-generated method stub
        super.startElement(uri, localName, qName, attributes);
        if(localName.equals(XML_CONTACT_NAME)){
            contactItem = new ContactName();
        }

        sb.setLength(0);

    }

    @Override
    public void characters(char[] ch, int start, int length){
        // TODO Auto-generated method stub
        try {
            super.characters(ch, start, length);
        } catch (SAXException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        sb.append(ch, start, length);
    }

    @Override
    public void endDocument() throws SAXException {
        // TODO Auto-generated method stub
        super.endDocument();
    }

    /**
     * where the real stuff happens
     */
    @Override
    public void endElement(String uri, String localName, String qName)
            throws SAXException {
        // TODO Auto-generated method stub
        //super.endElement(arg0, arg1, arg2);

        if(contactItem != null){
            if (localName.equalsIgnoreCase("title")) {
                contactItem.setUid(sb.toString());
                Log.d("handler", "setTitle = " + sb.toString());

            } else if (localName.equalsIgnoreCase("link")) {
                contactItem.setFullName(sb.toString());

            } else if (localName.equalsIgnoreCase("item")){
                Log.d("handler", "adding rss item");
                contactNameList.add(contactItem);
            }

            sb.setLength(0);
        }
}

Thanks in advance

You can look at this: http://stackoverflow.com/questions/247161/how-do-i-turn-a-string-into-a-stream-in-java — foret, Oct 11 '10 at 14:00

score 45 · Accepted Answer · edited Oct 11 '14 at 00:26

45

The SAXParser can read an InputSource.

An InputSource can take a Reader in its constructor

So, you can put parse XML string via a StringReader

new InputSource(new StringReader("... your xml here....")));

edited Oct 11 '14 at 00:26

Yiannis Tsimalis

739
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answered Oct 11 '10 at 14:03

Pierre

34,472
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score 3 · Answer 2 · answered Apr 24 '14 at 13:42

3

Try jcabi-xml (see this blog post) with a one-liner:

XML xml = new XMLDocument("<document>...</document>")

answered Apr 24 '14 at 13:42

yegor256

102,010
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score 1 · Answer 3 · answered Jul 15 '13 at 17:01

Your XML might be simple enough to parse manually using the DOM or SAX API, but I'd still suggest using an XML serialization API such as JAXB, XStream, or Simple instead because writing your own XML serialization/deserialization code is a drag.

Note that the XStream FAQ erroneously claims that you must use generated classes with JAXB:

How does XStream compare to JAXB (Java API for XML Binding)?

JAXB is a Java binding tool. It generates Java code from a schema and you are able to transform from those classes into XML matching the processed schema and back. Note, that you cannot use your own objects, you have to use what is generated.

It seems this was true was true at one time, but JAXB 2.0 no longer requires you to use Java classes generated from a schema.

If you go this route, be sure to check out the side-by-side comparisons of the serialization/marshalling APIs I've mentioned:

http://blog.bdoughan.com/2010/10/how-does-jaxb-compare-to-xstream.html http://blog.bdoughan.com/2010/10/how-does-jaxb-compare-to-simple.html

score 1 · Answer 4 · edited Jul 10 '15 at 16:16

Take a look at this: http://www.rgagnon.com/javadetails/java-0573.html

import javax.xml.parsers.*;
import org.xml.sax.InputSource;
import org.w3c.dom.*;
import java.io.*;

public class ParseXMLString {

  public static void main(String arg[]) {
     String xmlRecords =
      "<data>" +
      " <employee>" +
      "   <name>John</name>" +
      "   <title>Manager</title>" +
      " </employee>" +
      " <employee>" +
      "   <name>Sara</name>" +
      "   <title>Clerk</title>" +
      " </employee>" +
      "</data>";

    try {
        DocumentBuilderFactory dbf =
            DocumentBuilderFactory.newInstance();
        DocumentBuilder db = dbf.newDocumentBuilder();
        InputSource is = new InputSource();
        is.setCharacterStream(new StringReader(xmlRecords));

        Document doc = db.parse(is);
        NodeList nodes = doc.getElementsByTagName("employee");

        // iterate the employees
        for (int i = 0; i < nodes.getLength(); i++) {
           Element element = (Element) nodes.item(i);

           NodeList name = element.getElementsByTagName("name");
           Element line = (Element) name.item(0);
           System.out.println("Name: " + getCharacterDataFromElement(line));

           NodeList title = element.getElementsByTagName("title");
           line = (Element) title.item(0);
           System.out.println("Title: " + getCharacterDataFromElement(line));
        }
    }
    catch (Exception e) {
        e.printStackTrace();
    }
    /*
    output :
        Name: John
        Title: Manager
        Name: Sara
        Title: Clerk
    */    

  }

  public static String getCharacterDataFromElement(Element e) {
    Node child = e.getFirstChild();
    if (child instanceof CharacterData) {
       CharacterData cd = (CharacterData) child;
       return cd.getData();
    }
    return "?";
  }
}

parse an xml string in java?

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