你好，请教一下关于在json 中如何使用@JsonIgnore？

Question

json 中如何使用@JsonIgnore

Answer 1

public class JackJsonTest {public static void main(String[] args) throws IOException {User user = new User("abc", "id", 10);ObjectMapper objectMapper = new ObjectMapper();String json = objectMapper.writeValueAsString(user);System.out.println(json);User jsonUser = objectMapper.readValue(json, User.class);System.out.println(jsonUser.getAge());List<User> list = new ArrayList<User>();list.add(new User("abc1", "id1", 101));list.add(new User("abc2", "id2", 102));list.add(new User("abc3", "id3", 103));String listJson = objectMapper.writeValueAsString(list);System.out.println(listJson);List<User> beanList = objectMapper.readValue(listJson, new TypeReference<List<User>>() {});for (User jsonUserList : beanList) {System.out.println(jsonUserList);}}}class User {private String name;private String id;private Integer age;@JsonProperty(value = "aaa")public String getName() {return name;}@JsonProperty(value = "aaa")public void setName(String name) {this.name = name;}@JsonIgnorepublic String getId() {return id;}public void setId(String id) {this.id = id;}public Integer getAge() {return age;}public void setAge(Integer age) {this.age = age;}public User() {}public User(String name, String id, Integer age) {this.name = name;this.id = id;this.age = age;}@Overridepublic String toString() {return "User{" +"name='" + name + '\'' +", id='" + id + '\'' +", age=" + age +'}';}}

Answer 2

如果你使用的是：Newtonsoft.Json可以使用这个[JsonIgnore]标记，如：class I{public int Id { get; set; }public string Name { get; set; }[JsonIgnore]public char Sex { get; set; }}