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 Strin

Answer 1

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....")));

Answer 2

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

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

Answer 3

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 "?";
  }
}

Answer 4

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