为了确保在我的 RESTful Web 服务中来回发送的数据不是多余的,每个嵌套对象仅对其 ID 进行序列化(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用户的所有详细信息)。
序列化工作完美,产生这样的结果:
{"messageCreatorUser":"60d01602-c04d-4a3f-bbf2-132eb0ebbfc6","messageBody":"Body", ...}
问题:反序列化不会生成仅具有 ID 的嵌套对象。生成的反序列化嵌套对象为 null。
这里是前面提到的 Message
和 User
对象。序列化“策略”是从此处指定的第三个选项中使用的:How to serialize only the ID of a child with Jackson .
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,我希望仅使用 ID 为 60d01602-c04d-4a3f-bbf2-132eb0ebbfc6
创建一个嵌套的 User
对象
使用:
- Wildfly 10.0.Final:
- RESTEasy 3.0.15.Final
- RESTEasy Jackson 2 提供商 3.0.15.Final
- Jackson 2.6.3(注释、核心、数据绑定(bind)等)
为什么结果不同?
最佳答案
正如这里的答案( 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:
Message.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);
}
User.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)
,您需要在某处拥有 @JsonIdentityInfo(...)
。
关于java - 从 ID 到对象的 JSON 反序列化,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38443016/