从ID到Object的JSON反序列化

Question

为了确保在我的RESTful Web服务中来回发送的数据不是多余的,每个嵌套对象只将它的ID序列化(A Message的User创建者只有userId序列化,因为客户端和服务器都已经知道了所有的细节.用户).

序列化工作完美,产生:

{"messageCreatorUser":"60d01602-c04d-4a3f-bbf2-132eb0ebbfc6","messageBody":"Body", ...}

问题:反序列化不会生成仅包含其ID的嵌套对象.生成的反序列化嵌套对象为null.

这是前面提到的Message和User对象.从此处指定的第3个选项中使用序列化"策略": 如何仅使用Jackson序列化孩子的ID.

Message.java

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "messageId")
public class Message implements Serializable {

    // -- Hibernate definitions omitted --
    private UUID messageId;

    // -----------------------------------------------------------------
    // Here is what makes the serializer print out the ID specified on the class definition

    @JsonIdentityReference(alwaysAsId = true)

    // Here is my attempt to get the User class back when I deserialize

    @JsonDeserialize(as = User.class)
    @JsonSerialize(as = User.class)

    private User messageCreatorUser;
    // ------------------------------------------------------------------


    // -- more arbitrary properties --

    public Message() {
    }

    public Message(UUID messageId) {
        this.messageId = messageId;
    }

    public Message(String messageId) {
        this.messageId = UUID.fromString(messageId);
    }

    // -- getters and setters --

User.java

@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "userId")
@JsonInclude(JsonInclude.Include.NON_NULL)
public class User implements Serializable {

    private UUID userId;

    // -- other arbitrary properties -- 

    public User() {
    }

    public User(UUID userId) {
        this.userId = userId;
    }

    public User(String userId) {
        this.userId = UUID.fromString(userId);
    }

    // -- getters and setters --

预期的反序列化对象:

Message object =
     String messageBody = "Body";
     User messageCreatorUser =
         UUID userId = 60d01602-c04d-4a3f-bbf2-132eb0ebbfc6;

实际反序列化对象:

Message object =
     String messageBody = "Body";
     User messageCreatorUser = null;

就像我说的那样,我希望User只用ID来创建一个嵌套对象60d01602-c04d-4a3f-bbf2-132eb0ebbfc6

使用:

• Wildfly 10.0.Final:
  • RESTEasy 3.0.15.Final
  • RESTEasy Jackson 2 Provider 3.0.15.Final
    • 杰克逊2.6.3(注释,核心,数据绑定等)

为什么结果不同？

Answer 1

正如此处的答案（Jackson deserialize JsonIdentityReference (alwaysAsId = true)）所述，使用 setter 反序列化器效果很好，并且不需要令人讨厌的自定义反序列化器：

消息.java

    @JsonInclude(JsonInclude.Include.NON_NULL)
    @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "messageId")
    public class Message implements Serializable {

        private UUID messageId;

        @JsonIdentityReference(alwaysAsId = true)
        private User messageCreatorUser;

        // -- other fields and such --

        public User getMessageCreatorUser() {
            return messageCreatorUser;
        }

        public void setMessageCreatorUser(User messageCreatorUser) {
            this.messageCreatorUser = messageCreatorUser;
        }

        @JsonProperty("messageCreatorUser")
        public void setMessageCreatorUser(String userId) {
            this.messageCreatorUser = new User(userId);
        }

用户.java

    @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "userId")
    public class User implements Serializable {

        private UUID userId;

        // -- other fields --

        public User(String userId) {
            this.userId = UUID.fromString(userId);
        }

    }

请注意，为了使用@JsonIdentityReference(alwaysAsId = true)，您需要有@JsonIdentityInfo(...)某个地方。