如何将一个复杂的Java对象序列化为xml 使用XStream库的简单方法

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<dependency>
    <groupId>com.thoughtworks.xstream</groupId>
    <artifactId>xstream</artifactId>
    <version>1.4.20</version>
</dependency>
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class Address {
    private String city;
    private String zipCode;

    public Address(String city, String zipCode) {
        this.city = city;
        this.zipCode = zipCode;
    }
    // toString 省略
}
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class User {
    private String name;
    private int age;
    private List<Address> addresses;

    public User(String name, int age, List<Address> addresses) {
        this.name = name;
        this.age = age;
        this.addresses = addresses;
    }
}
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import com.thoughtworks.xstream.XStream;
import java.util.Arrays;

public class XmlSerializationExample {
    public static void main(String[] args) {
        // 构建复杂对象
        List<Address> addresses = Arrays.asList(
            new Address("北京", "100001"),
            new Address("上海", "200001")
        );
        User user = new User("张三", 30, addresses);

        // 使用XStream序列化
        XStream xstream = new XStream();
        xstream.alias("user", User.class);
        xstream.alias("address", Address.class);

        String xml = xstream.toXML(user);
        System.out.println(xml);
    }
}
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<user>
  <name>张三</name>
  <age>30</age>
  <addresses>
    <address>
      <city>北京</city>
      <zipCode>100001</zipCode>
    </address>
    <address>
      <city>上海</city>
      <zipCode>200001</zipCode>
    </address>
  </addresses>
</user>
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User restoredUser = (User) xstream.fromXML(xml);
System.out.println(restoredUser.getName()); // 输出：张三
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