Java读写xml（对简单关系，无需中间模型）
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输出结果：
[y1, x2]
[y2, x1, z3]-------------------------------------------
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stus/stu类中的List也可以初始化，于是代码如下：
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import java.io.File;import java.util.ArrayList;import java.util.List;import javax.xml.bind.JAXBContext;import javax.xml.bind.JAXBException;import javax.xml.bind.Marshaller;import javax.xml.bind.Unmarshaller;import javax.xml.bind.annotation.XmlRootElement;public class test2 {public static void main(String[] args) {write();read();}public static void write() {try {JAXBContext jc = JAXBContext.newInstance(stus.class);Marshaller u = jc.createMarshaller();stus stus = new stus();// stus.stu = new ArrayList<stu>();stu stu = new stu();// stu.pet = new ArrayList<String>();stu.pet.add("y1");stu.pet.add("x2");stus.stu.add(stu);stu = new stu();// stu.pet = new ArrayList<String>();stu.pet.add("y2");stu.pet.add("x1");stu.pet.add("z3");stus.stu.add(stu);u.marshal(stus, new File("d:\\2.xml"));} catch (JAXBException e) {e.printStackTrace();}}public static void read() {try {JAXBContext jc = JAXBContext.newInstance(stus.class);Unmarshaller u = jc.createUnmarshaller();stus stus = (stus) u.unmarshal(new File("d:\\2.xml"));for (stu stu : stus.stu) {System.out.println(stu.pet);}} catch (JAXBException e) {e.printStackTrace();}}}@XmlRootElementclass stus {public List<stu> stu = new ArrayList<stu>();}class stu {public List<String> pet = new ArrayList<String>();}?
