我正在处理的项目使用JAXB参考实现,也就是说这些类都来自com.sun.xml.bind.v2.*
包。
我有一个名为User
的类:
package com.example;
import javax.xml.bind.annotation.XmlRootElement;
@XmlRootElement(name = "user")
public class User {
private String email;
private String password;
public User() {
}
public User(String email, String password) {
this.email = email;
this.password = password;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
}
我想使用JAXB marshaller将User
对象转换成JSON格式:
@Test
public void serializeObjectToJson() throws JsonProcessingException, JAXBException {
User user = new User("user@example.com", "mySecret");
JAXBContext jaxbContext = JAXBContext.newInstance(User.class);
Marshaller marshaller = jaxbContext.createMarshaller();
StringWriter sw = new StringWriter();
marshaller.marshal(user, sw);
assertEquals( "{\"email\":\"user@example.com\", \"password\":\"mySecret\"}", sw.toString() );
}
序列化数据是以XML格式而非JSON格式进行的。 如何指示JAXB参考实现输出JSON?