请教java如何操作xml文件

Question

求助

Answer 1

public class Sax extends DefaultHandler {    /**
     * @param args
     * @throws SAXException 
     * @throws ParserConfigurationException 
     * @throws IOException 
     */
    public static void main(String[] args) throws ParserConfigurationException,
            SAXException, IOException {        SAXParserFactory spf = SAXParserFactory.newInstance();
        spf.setValidating(false);
        spf.setFeature("http://xml.org/sax/features/namespace-prefixes", true);
        spf.setFeature("http://xml.org/sax/features/namespaces", true);
        SAXParser sp = spf.newSAXParser();
        XMLReader xr = sp.getXMLReader();        xr.setContentHandler(new Sax());
        xr.setErrorHandler(new Sax());        InputSource is = new InputSource("c:\\sax.xml");
        xr.parse(is);
    }
    public void startDocument() throws SAXException {        System.out.println("start to read xml");
    }
    public void endDocument() throws SAXException {        System.out.println("end of reading xml");
    }
    public void startElement(String uri, String localName, String qName,
            Attributes attributes) throws SAXException {//        System.out.println("uri----" + uri);
//        System.out.println("localName----" + localName);
//        System.out.println("qName----" + qName);
//        Method[] methods = attributes.getClass().getMethods();
//        for (int i = 0; i < methods.length; i++) {
//            Method method = methods[i];
//            String name = method.getName();
//            if (name.startsWith("get")) {
//                Class[] param = method.getParameterTypes();
//                if (param.length == 1 && param[0].isPrimitive()) {
//                    try {
//                        for (int j = 0; j < attributes.getLength(); j++) {
//                            String info = name
//                                    + "----"
//                                    + method.invoke(attributes,
//                                            new Object[] { new Integer(j) });
//                            System.out.println(info);
//                        }
//                    } catch (IllegalArgumentException e) {
//                        // TODO Auto-generated catch block
//                        e.printStackTrace();
//                    } catch (IllegalAccessException e) {
//                        // TODO Auto-generated catch block
//                        e.printStackTrace();
//                    } catch (InvocationTargetException e) {
//                        // TODO Auto-generated catch block
//                        e.printStackTrace();
//                    }
//                }
//            }
//        }
//        System.out.println("----");
    }
    public void processingInstruction(String target, String data)
            throws SAXException {        System.out.println("target----" + target);
        System.out.println("data----" + data);
    }}

Answer 2

BufferReader in=new BufferReader(new FileReader(xml文件路径));
DOMParser parser=new DOMParser();
parser.parse(new InputSource(in));
Document doc=parser.getDocument();
然后使用doc.getElementsByTagName("*")获得所有xml中的元素；

Answer 3

把dom4j放到环境变量下就可以了