如何从字符串中加载XML中的org.w3c.dom.Document？

Question

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, java.io.IOException {
    return loadXMLFrom(new java.io.ByteArrayInputStream(xml.getBytes()));
}

public static org.w3c.dom.Document loadXMLFrom(java.io.InputStream is) 
    throws org.xml.sax.SAXException, java.io.IOException {
    javax.xml.parsers.DocumentBuilderFactory factory =
        javax.xml.parsers.DocumentBuilderFactory.newInstance();
    factory.setNamespaceAware(true);
    javax.xml.parsers.DocumentBuilder builder = null;
    try {
        builder = factory.newDocumentBuilder();
    }
    catch (javax.xml.parsers.ParserConfigurationException ex) {
    }  
    org.w3c.dom.Document doc = builder.parse(is);
    is.close();
    return doc;
}

Answer 1

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 java.io.StringReader;
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.

Answer 2

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;
import java.io.ByteArrayInputStream;

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

    factory.setNamespaceAware(true);
    DocumentBuilder builder = factory.newDocumentBuilder();

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

Answer 3

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();
        factory.setNamespaceAware(true);
        DocumentBuilder builder = null;
        builder = factory.newDocumentBuilder();
        Document doc = builder.parse(is);
        return doc;
    }

Answer 4

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;
import javax.xml.transform.stream.StreamSource;

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();
}