如何用java封装解析一个多层次的XML文件，急求大神帮助

Question

如题所述

Answer 1

下面提供了XML解析成实体类，以及实体类转换成xml的方法。
public class SwitchXML {

private XStream xStream = new XStream();

private String xmlPath = "D:/data.xml";

// 1. obj -> xml(object类型转换为xml类型)
public String printXML(User user) {
System.out.println("obj -> xml");
String xml = xStream.toXML(user);
System.out.println(xml);
return xml;
}

// 2. xml->obj(xml类型转换为object类型，并打印)
public void printObj(String xml) {
System.out.println("xml -> obj");
User u = (User) xStream.fromXML(xml);
System.out.println(u.getUserName() + " " + u.getPassWord() + " " + u.getRegisteredTime());
}

// 3. 将object类型转换为xml类型，并写入XML文件(其他格式也可以，比如txt文件)
public void writerXML(User user) {
try {
FileOutputStream fs = new FileOutputStream(xmlPath);
xStream.toXML(user, fs);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}

// 4. 读取XML文件，加载进相应Object类型
public void readerXML() {
User user = new User();
FileInputStream fis = null;
try {
fis = new FileInputStream(xmlPath);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
xStream.fromXML(fis, user);
// 打印对象信息
System.out.println(user.toString());
// 打印对象的属性值
System.out.println(user.getUserName() + "-" + user.getPassWord() + "-"
+ user.getRegisteredTime());
}

public static void main(String[] args) {
// 用户名
String username = "admin";
// 密码
String password = "admin";
// 注册时间
Date now = new Date();
SimpleDateFormat format = new SimpleDateFormat("yyyyMMddHHmmss");
String registeredtime = format.format(now);

User user = new User(username, password, registeredtime);
SwitchXML switchXML = new SwitchXML();

// 1. object类型转换为xml类型，在控制台打印
String xml = switchXML.printXML(user);
System.out.println("---------------------");
// 2. xml类型转换为object类型，在控制台打印
switchXML.printObj(xml);
System.out.println("---------------------");
// 3. 将object类型转换为xml类型，并写入XML文件
switchXML.writerXML(user);
System.out.println("---------------------");
// 4. 读取XML文件，加载进相应Object类型
switchXML.readerXML();

}