如何在Java中解析包含XML的String并检索根节点的值？

Question

我有一个包含的String形式的XML

<message>HELLO!</message> 

如何获取字符串"Hello!" 来自XML？这应该是非常容易的,但我迷失了.XML不在doc中,它只是一个String.

Answer 1

使用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
}

使用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 
}

使用JAXP接口:

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

您还可以使用基本JRE提供的工具:

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

您可以使用JAXB执行此操作(Java SE 6中包含一个实现).

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());
    }

}

产量

HELLO!

Answer 4

上述答案之一指出将 XML 字符串转换为不需要的字节。相反，您可以使用 InputSource和提供它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());