【问题标题】:Cannot deserialize generic class hierarchy using Jackson无法使用 Jackson 反序列化泛型类层次结构
【发布时间】:2019-04-09 08:03:33
【问题描述】:

例如,我从外部供应商那里收到了这个 JSON(payload 可以是可变的):

{
  "payload": {
    "enrolledAt": "2018-11-05T00:00:00-05:00",
    "userId": "99c7ff5c-2c4e-423f-abeb-2e5f3709a42a"
  },
  "requestId": "80517bb8-2a95-4f15-9a73-fcf3752a1147",
  "eventType": "event.success",
  "createdAt": "2018-11-05T16:55:13.762-05:00"
}

我正在尝试使用此类对它们进行建模:

public final class Notification<T extends AbstractModel> {
  @JsonProperty("requestId")
  private String requestId;

  @JsonProperty("eventType")
  private String eventType;

  @JsonProperty("createdAt")
  private ZonedDateTime createdAt;

  private T payload;

  @JsonCreator
  public Notification(@JsonProperty("payload") T payload) {
    requestId = UUID.randomUUID().toString();
    eventType = payload.getType();
    createdAt = ZonedDateTime.now();
    this.payload = payload;
  }

  // getters
}

...然后拥有这些可能的(通用)类型:

public abstract class AbstractModel {
  private String userId;

  private Type type;

  @JsonCreator
  AbstractModel(@JsonProperty("companyUserId") String userId, @JsonProperty("type") Type type) {
    this.userId = userId;
    this.type = type;
  }

  // getters

  public enum Type {
    CANCEL("event.cancel"),
    SUCCESS("event.success");

    private final String value;

    Type(String value) {
      this.value = value;
    }

    public String getValue() { return value; }
  }
}

public final class Success extends AbstractModel {
  private ZonedDateTime enrolledAt;

  @JsonCreator
  public Success(String userId, @JsonProperty("enrolledAt") ZonedDateTime enrolledAt) {
    super(userId, Type.SUCCESS);
    this.enrolledAt = enrolledAt;
  }

  // getters
}

public final class Cancel extends AbstractModel {
  private ZonedDateTime cancelledAt;

  private String reason;

  @JsonCreator
  public Cancel(String userId, @JsonProperty("cancelledAt") ZonedDateTime cancelledAt,
      @JsonProperty("reason") String reason) {
    super(userId, Type.CANCEL);
    this.cancelledAt = cancelledAt;
    this.reason = reason;
  }

  // getters
}

该应用程序基于 Spring Boot,因此我将 JSON 反序列化:

@Component
public final class NotificationMapper {    
  private ObjectMapper mapper;

  public NotificationMapper(final ObjectMapper mapper) {
    this.mapper = mapper;
  }

  public Optional<Notification<? extends AbstractModel>> deserializeFrom(final String thiz) {
    try {
      return Optional.of(mapper.readValue(thiz, new NotificationTypeReference()));
    } catch (final Exception e) { /* log errors here */ }
    return Optional.empty();
  }

  private static final class NotificationTypeReference extends TypeReference<Notification<? extends AbstractModel>> { }
}

...但最终因为我在这里发布这个,Jackson 到目前为止不喜欢其中的任何一个。我尝试了几种方法,例如:JsonTypeInfoJsonSubTypes,但我无法更改 JSON 输入。

有人吗?有什么线索吗?

【问题讨论】:

    标签: java spring-boot jackson jackson-databind


    【解决方案1】:

    我们最终在 JSON 中添加了另一个键/值对 - 确实对合约进行了一些修改以适应该“私有”键/值对。

    无论如何,如果有人遇到同样的问题,这是该方法的解决方案:

    ...
    import com.fasterxml.jackson.annotation.JsonIgnore;
    import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
    import com.fasterxml.jackson.annotation.JsonSubTypes;
    import com.fasterxml.jackson.annotation.JsonTypeInfo;
    
    @JsonIgnoreProperties("_type")
    @JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "_type")
    @JsonSubTypes({
      @JsonSubTypes.Type(value = Cancel.class, name = "_type"),
      @JsonSubTypes.Type(value = Fail.class, name = "_type"),
      @JsonSubTypes.Type(value = Success.class, name = "_type"),
    })
    public abstract class AbstractModel {
      private String userId;
    
      private Type type;
    
      AbstractModel() { }
    
      AbstractModel(final String userId, final Type type) {
        this.userId = userId;
        this.type = type;
      }
    
      // getters, toString, etc.
    
      public enum Type {
        CANCEL("event.cancelled"),
        FAIL("event.failed"),
        SUCCESS("event.success");
    
        private final String value;
    
        Type(String value) {
          this.value = value;
        }
    
        public String getValue() { return value; }
      }
    }
    

    【讨论】:

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