Java 将JSON映射到类不起作用
我有一个包含其他类的其他属性的类,当我尝试从json转换到我的类时,会显示一个错误 这是我的班级:Java 将JSON映射到类不起作用,java,json,mongodb,jackson,jongo,Java,Json,Mongodb,Jackson,Jongo,我有一个包含其他类的其他属性的类,当我尝试从json转换到我的类时,会显示一个错误 这是我的班级: import org.jongo.marshall.jackson.oid.MongoObjectId; import org.json.JSONObject; import java.util.List; public class BusinessTravelDTO { @MongoObjectId private String id; private String trave
import org.jongo.marshall.jackson.oid.MongoObjectId;
import org.json.JSONObject;
import java.util.List;
public class BusinessTravelDTO {
@MongoObjectId
private String id;
private String travelerId;
private BusinessTravelStatus status;
List<FlightDTO> flights;
List<HotelDTO> hotels;
List<CarDTO> cars;
public BusinessTravelDTO() { }
public BusinessTravelDTO(JSONObject data) {
this.travelerId = data.getString("travelerId");
this.status = BusinessTravelStatus.valueOf(data.getString("status"));
this.flights = HandlerUtil.getInputFlights(data.getJSONArray("flights"));
this.hotels = HandlerUtil.getInputHotels(data.getJSONArray("hotels"));
this.cars = HandlerUtil.getInputCars(data.getJSONArray("cars"));
}
public JSONObject toJson() {
return new JSONObject()
.put("id", this.id)
.put("travelerId", this.travelerId)
.put("status", this.status)
.put("flights", this.flights)
.put("hotels", this.hotels)
.put("cars", this.cars);
}
以下是我收到的错误:
"error": "org.jongo.marshall.MarshallingException: Unable to unmarshall result to class path.data.BusinessTravelDTO from content { \"_id\" : { \"$oid\" : \"59d6905411d58632fd5bd8a5\"} , \"travelerId\"
在jongo文档中,指定类应该有一个空构造函数。。。我有两个构造函数,我也尝试过使用@JsonCreator,但没有成功…:(你知道为什么它不能转换吗?它可能与businessTravelTo中的字段有关,比如List CarDTO for ex?我最终找到了解决方案 FlightDTO、HotelDTO、CarDTO等所有类中都需要一个空构造函数,我应该按照以下方式重写toJson方法:
public JSONObject toJson() {
JSONObject obj = new JSONObject().put("id", this.id).put("travelerId", this.travelerId).put("status", this.status);
if (flights != null) {
JSONArray flightArray = new JSONArray();
for (int i = 0; i < flights.size(); ++i) {
flightArray.put(flights.get(i).toJson());
}
obj.put("flights", flightArray);
}
if (hotels != null) {
JSONArray hotelArray = new JSONArray();
for (int i = 0; i < hotels.size(); ++i) {
hotelArray.put(hotels.get(i).toJson());
}
obj.put("hotels", hotelArray);
}
if (cars != null) {
JSONArray carArray = new JSONArray();
for (int i = 0; i < cars.size(); ++i) {
carArray.put(cars.get(i).toJson());
}
obj.put("cars", carArray);
}
return obj;
}
现在它工作得很好!:)JSON文本似乎有一个带有
$oid
字段的某个对象类型的\u id
字段,而您的BusinessTravelDTO
有一个id
类型的字段。你以为那会怎样不过我有点困惑,因为我看不出您在代码中的什么地方进行编组。向我们展示acceptBusinessTravel()
方法的意义是什么?@Andreas您可能错过了@MongoObjectId,这个选项允许您在类中使用mongo对象id,但如果您不使用它,那么“\u id”无论如何都是在幕后创建的…:-|我看到了,但这和JSON有什么关系?
public JSONObject toJson() {
JSONObject obj = new JSONObject().put("id", this.id).put("travelerId", this.travelerId).put("status", this.status);
if (flights != null) {
JSONArray flightArray = new JSONArray();
for (int i = 0; i < flights.size(); ++i) {
flightArray.put(flights.get(i).toJson());
}
obj.put("flights", flightArray);
}
if (hotels != null) {
JSONArray hotelArray = new JSONArray();
for (int i = 0; i < hotels.size(); ++i) {
hotelArray.put(hotels.get(i).toJson());
}
obj.put("hotels", hotelArray);
}
if (cars != null) {
JSONArray carArray = new JSONArray();
for (int i = 0; i < cars.size(); ++i) {
carArray.put(cars.get(i).toJson());
}
obj.put("cars", carArray);
}
return obj;
}
public class FlightDTO {
@MongoObjectId
private String id;
private String departure;
private String arrival;
private String airline;
private Double price;
public FlightDTO() {
}
public FlightDTO(JSONObject data) {
this.departure = data.getString("departure");
this.arrival = data.getString("arrival");
this.airline = data.getString("airline");
this.price = data.getDouble("price");
}
public JSONObject toJson() {
return new JSONObject()
.put("id", this.id)
.put("departure", this.departure)
.put("arrival", this.arrival)
.put("airline", this.airline)
.put("price", this.price);
}
}