@JsonRootName 不起作用

Question

我有这个简单的 Java 实体，我需要将其设为 JSon 输出，我可以通过 e Web 服务访问它。

@Entity
@JsonRootName(value = "flights")
public class Flight implements Serializable {

@Transient
private static final long serialVersionUID = 1L;

public Flight() {
    super();
}

public Flight(FlightDestination destinationFrom, FlightDestination destinationTo, Integer flightPrice, Date date,
        Airplane airplaneDetail) {
    super();
    this.destinationFrom = destinationFrom;
    this.destinationTo = destinationTo;
    this.flightPrice = flightPrice;
    this.date = date;
    this.airplaneDetail = airplaneDetail;
}

public Flight(FlightDestination destinationFrom, FlightDestination destinationTo, Integer flightPrice, Date date) {
    super();
    this.destinationFrom = destinationFrom;
    this.destinationTo = destinationTo;
    this.flightPrice = flightPrice;
    this.date = date;
}

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;

@Enumerated(EnumType.STRING)
private FlightDestination destinationFrom;

@Enumerated(EnumType.STRING)
private FlightDestination destinationTo;

private Integer flightPrice;

@Temporal(TemporalType.DATE)
private Date date;

@OneToOne(cascade = { CascadeType.PERSIST, CascadeType.REMOVE })
@JoinColumn(name = "airplane_fk")
private Airplane airplaneDetail;}

我添加了@JsonRootName，但我仍然以这种方式获得我的 json 输出：

    [  
      {   

      },

      { 

      }
   ]

我还有什么要添加到我的实体，所以最终得到这种输出：

    {
     "flights":

     [  

      {   

      },

      { 

      }
    ]
   }

Answer 1

如果你想使用@JsonRootName(value = "flights")，你必须在ObjectMapper上设置适当的功能

ObjectMapper mapper = new ObjectMapper();
mapper.enable(DeserializationFeature.UNWRAP_ROOT_VALUE); 
mapper.enable(SerializationFeature.WRAP_ROOT_VALUE);

但是，对于List<Flight>，这将产生

[  
  {"flights": {}},
  {"flights": {}},
  {"flights": {}},
]

所以你可能必须创建包装对象：

public class FlightList {
    @JsonProperty(value = "flights")
    private ArrayList<Flight> flights;
}

这个FlightList 将有{"flights":[{ }, { }]} 输出json

Answer 2

您可以使用以下注解来使用 Jackson 来制作包装器；

...
@JsonTypeInfo(include = JsonTypeInfo.As.WRAPPER_OBJECT, use = JsonTypeInfo.Id.NAME)
@JsonTypeName("user")
public class LoginResponse {
  private String name;
  private String email;
}

回复将是；

{
  "user": {
    "name": "Lucienne",
    "username": "asenocakUser"
  }
}