Java中fastjson的toJSONString结果为空{}

public class User {
    @JSONField(name = "ID")
    private String id;
    private String name;
    private String age;
    @JSONField(name = "LIKE_FOOD")
    private String likeFood;
    private Job job;

    public User(String id, String name, String age, String likeFood, Job job) {
        this.id = id;
        this.name = name;
        this.age = age;
        this.likeFood = likeFood;
        this.job = job;
    }

    @Override
    public String toString(){
        return JSONObject.toJSONString(this,true);
    }
}

public class JsonTest {
    public static void main(String[] args) {
       User user = new User("1", "yang", "18", "apple", new Job("bank", "18000"));
       System.out.println(user.toString());
    }
}

{
	"ID":"1",
	"LIKE_FOOD":"apple",
	"age":"18",
	"job":{
		"jobName":"bank",
		"salary":"18000"
	},
	"name":"yang"
}