How do I load an org.w3c.dom.Document from XML in a string?


I have a complete XML document in a string and would like a Document object. Google turns up all sorts of garbage. What is the simplest solution? (In Java 1.5)

Solution Thanks to Matt McMinn, I have settled on this implementation. It has the right level of input flexibility and exception granularity for me. (It's good to know if the error came from malformed XML - SAXException - or just bad IO - IOException.)

public static org.w3c.dom.Document loadXMLFrom(String xml)
    throws org.xml.sax.SAXException, {
    return loadXMLFrom(new;

public static org.w3c.dom.Document loadXMLFrom( is) 
    throws org.xml.sax.SAXException, {
    javax.xml.parsers.DocumentBuilderFactory factory =
    javax.xml.parsers.DocumentBuilder builder = null;
    try {
        builder = factory.newDocumentBuilder();
    catch (javax.xml.parsers.ParserConfigurationException ex) {
    org.w3c.dom.Document doc = builder.parse(is);
    return doc;

This question is tagged with java xml document w3c

~ Asked on 2008-08-28 20:03:19

4 Answers


This works for me in Java 1.5 - I stripped out specific exceptions for readability.

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.w3c.dom.Document;

public Document loadXMLFromString(String xml) throws Exception
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();

    DocumentBuilder builder = factory.newDocumentBuilder();

    return builder.parse(new ByteArrayInputStream(xml.getBytes()));

~ Answered on 2008-08-28 20:11:00


Whoa there!

There's a potentially serious problem with this code, because it ignores the character encoding specified in the String (which is UTF-8 by default). When you call String.getBytes() the platform default encoding is used to encode Unicode characters to bytes. So, the parser may think it's getting UTF-8 data when in fact it's getting EBCDIC or something… not pretty!

Instead, use the parse method that takes an InputSource, which can be constructed with a Reader, like this:

import org.xml.sax.InputSource;
        return builder.parse(new InputSource(new StringReader(xml)));

It may not seem like a big deal, but ignorance of character encoding issues leads to insidious code rot akin to y2k.

~ Answered on 2008-08-28 23:16:54


Just had a similar problem, except i needed a NodeList and not a Document, here's what I came up with. It's mostly the same solution as before, augmented to get the root element down as a NodeList and using erickson's suggestion of using an InputSource instead for character encoding issues.

private String DOC_ROOT="root";
String xml=getXmlString();
Document xmlDoc=loadXMLFrom(xml);
Element template=xmlDoc.getDocumentElement();
NodeList nodes=xmlDoc.getElementsByTagName(DOC_ROOT);

public static Document loadXMLFrom(String xml) throws Exception {
        InputSource is= new InputSource(new StringReader(xml));
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = null;
        builder = factory.newDocumentBuilder();
        Document doc = builder.parse(is);
        return doc;

~ Answered on 2008-09-05 15:57:59


To manipulate XML in Java, I always tend to use the Transformer API:

import javax.xml.transform.Source;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMResult;

public static Document loadXMLFrom(String xml) throws TransformerException {
    Source source = new StreamSource(new StringReader(xml));
    DOMResult result = new DOMResult();
    TransformerFactory.newInstance().newTransformer().transform(source , result);
    return (Document) result.getNode();

~ Answered on 2014-08-05 11:01:20

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