如何在Java中解析包含XML的字符串并检索根节点的值？

Question

I have XML in the form of a String that contains

<message>HELLO!</message> 

How can I get the String "Hello!" from the XML? It should be ridiculously easy but I am lost. The XML isn't in a doc, it is simply a String.

Answer 1

Using JDOM :

String xml = "<message>HELLO!</message>";
org.jdom.input.SAXBuilder saxBuilder = new SAXBuilder();
try {
    org.jdom.Document doc = saxBuilder.build(new StringReader(xml));
    String message = doc.getRootElement().getText();
    System.out.println(message);
} catch (JDOMException e) {
    // handle JDOMException
} catch (IOException e) {
    // handle IOException
}

Using the Xerces DOMParser :

String xml = "<message>HELLO!</message>";
DOMParser parser = new DOMParser();
try {
    parser.parse(new InputSource(new java.io.StringReader(xml)));
    Document doc = parser.getDocument();
    String message = doc.getDocumentElement().getTextContent();
    System.out.println(message);
} catch (SAXException e) {
    // handle SAXException 
} catch (IOException e) {
    // handle IOException 
}

Using the JAXP interfaces:

String xml = "<message>HELLO!</message>";
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = null;
try {
    db = dbf.newDocumentBuilder();
    InputSource is = new InputSource();
    is.setCharacterStream(new StringReader(xml));
    try {
        Document doc = db.parse(is);
        String message = doc.getDocumentElement().getTextContent();
        System.out.println(message);
    } catch (SAXException e) {
        // handle SAXException
    } catch (IOException e) {
        // handle IOException
    }
} catch (ParserConfigurationException e1) {
    // handle ParserConfigurationException
}

Answer 2

You could also use tools provided by the base JRE:

String msg = "<message>HELLO!</message>";
DocumentBuilder newDocumentBuilder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document parse = newDocumentBuilder.parse(new ByteArrayInputStream(msg.getBytes()));
System.out.println(parse.getFirstChild().getTextContent());

Answer 3

You could do this with JAXB (an implementation is included in Java SE 6).

import java.io.StringReader;
import javax.xml.bind.*;
import javax.xml.transform.stream.StreamSource;

public class Demo {

    public static void main(String[] args) throws Exception {
        String xmlString = "<message>HELLO!</message> ";
        JAXBContext jc = JAXBContext.newInstance(String.class);
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        StreamSource xmlSource = new StreamSource(new StringReader(xmlString));
        JAXBElement<String> je = unmarshaller.unmarshal(xmlSource, String.class);
        System.out.println(je.getValue());
    }

}

Output

HELLO!

Answer 4

One of the above answer states to convert XML String to bytes which is not needed. Instead you can can use InputSource and supply it with StringReader .

String xmlStr = "<message>HELLO!</message>";
DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document doc = db.parse(new InputSource(new StringReader(xmlStr)));
System.out.println(doc.getFirstChild().getNodeValue());

Answer 5

There is doing XML reading right, or doing the dodgy just to get by. Doing it right would be using proper document parsing.

Or... dodgy would be using custom text parsing with either wisuzu's response or using regular expressions with matchers.

Answer 6

I think you would be look at String class, there are multiple ways to do it. What about substring(int,int) and indexOf(int) lastIndexOf(int) ?