管理对象 java 列表中的一个对象

Question

我有对象列表，该对象列表多次显示相同的对象，但所有重复对象的颜色字段不同。我想更改响应。

如果有人对此有解决方案，请创建通用方法并传递此列表并根据我的期望响应隐藏。

1.实际反应

[
    {
      "id": 21,
      "startDate": "2020-07-21",
      "startTime": "10:28:34+05:30",
      "endDate": "2020-07-21",
      "endTime": "19:28:34+05:30",
      "eventName": "Stretegic Planing for Business Success",
      "location": "America Phoenix",
      "city": "San Jose de Gracia",
      "eventImg": "Image Url",
      "passport": false,
      "passportDiscountPercent": 5,
      "featured": true,
      "myFavourite": true,
      "color": "Yellow"
    },
    {
      "id": 21,
      "startDate": "2020-07-21",
      "startTime": "10:28:34+05:30",
      "endDate": "2020-07-21",
      "endTime": "19:28:34+05:30",
      "eventName": "Stretegic Planing for Business Success",
      "location": "America Phoenix",
      "city": "San Jose de Gracia",
      "eventImg": "Image Url",
      "passport": false,
      "passportDiscountPercent": 5,
      "featured": true,
      "myFavourite": true,
      "color": "Green"
    },
    {
      "id": 76,
      "startDate": "2020-07-21",
      "startTime": "10:28:34+05:30",
      "endDate": "2020-07-21",
      "endTime": "19:28:34+05:30",
      "eventName": "Stretegic Planing for Business Success",
      "location": "string",
      "city": "Villa Juarez",
      "eventImg": "Imgage url",
      "passport": false,
      "passportDiscountPercent": 5,
      "featured": false,
      "myFavourite": true,
      "color": "Red"
    },
    {
      "id": 76,
      "startDate": "2020-07-21",
      "startTime": "10:28:34+05:30",
      "endDate": "2020-07-21",
      "endTime": "19:28:34+05:30",
      "eventName": "Stretegic Planing for Business Success",
      "location": "string",
      "city": "Villa Juarez",
      "eventImg": "Imgage url",
      "passport": false,
      "passportDiscountPercent": 5,
      "featured": false,
      "myFavourite": true,
      "color": "Black"
    }
]    

2.期待回应

[
    {
      "id": 21,
      "startDate": "2020-07-21",
      "startTime": "10:28:34+05:30",
      "endDate": "2020-07-21",
      "endTime": "19:28:34+05:30",
      "eventName": "Stretegic Planing for Business Success",
      "location": "America Phoenix",
      "city": "San Jose de Gracia",
      "eventImg": "Image Url",
      "passport": false,
      "passportDiscountPercent": 5,
      "featured": true,
      "myFavourite": true,
      "eventType": [
            {
              "color": "Yellow"
            },
            {
              "color": "Green"
            }
      ]
    },
    {
      "id": 76,
      "startDate": "2020-07-21",
      "startTime": "10:28:34+05:30",
      "endDate": "2020-07-21",
      "endTime": "19:28:34+05:30",
      "eventName": "Stretegic Planing for Business Success",
      "location": "string",
      "city": "Villa Juarez",
      "eventImg": "Imgage url",
      "passport": false,
      "passportDiscountPercent": 5,
      "featured": false,
      "myFavourite": true,
      "eventType": [
            {
              "color": "Red"
            },
            {
              "color": "Black"
            }
      ]
    }   
]    

Answer 1

第 1 步：创建架构类

初始响应模型：

public class InitialModel {
            String id;
            String color;
}

最终/转换后的响应模型：

public class Model {
        String id;
        List<Color> eventType = new ArrayList<>();

        class Color {
            String color;

            public Color(String color) {
                this.color = color;
            }
        }
    }

第 2 步：迭代初始响应并将其转换为最终响应

List<Model> modelList = new ArrayList<>();
List<InitialModel> initialModelList = new Gson().fromJson(jsonString, new TypeToken<List<Model>>(){}.getType());

for (InitialModel initialModel : initialModelList) {

Model tempModel = null;

// Check if model with same id already exists
for(Model model : modelList){
    if(initialModel.id.equals(model.id)){
        tempModel = model;
        break;
    }
}

// Create new model if id doesn't exists
if(tempModel==null) {
    tempModel = new Model();
    tempModel.id = initialModel.id;
    modelList.add(tempModel);
}
tempModel.eventType.add(new Color(initialModel.color));


}

结果 Json：

String resultJson = new Gson().toJson(modelList);

1.我已经添加了您的代码，但出现了一些错误

private List<EventCustomDTO> commonListView(List<EventDtoClass> eventDtoClassList) {
        List<EventCustomDTO> eventCustomDTOList = new ArrayList<>();
        List<EventDtoClass> eventDtoClasses = new Gson().fromJson(eventDtoClassList.toString(), new TypeToken<List<EventCustomDTO>>(){}.getType());
        for (EventDtoClass eventDtoClass : eventDtoClasses) {
            EventCustomDTO eventCustomDTO = null;
            for(EventCustomDTO model : eventCustomDTOList){
                if(eventDtoClass.getId().equals(model.getId())){
                    eventCustomDTO = model;
                    break;
                }
            }
            if(eventCustomDTO == null) {
                eventCustomDTO = new EventCustomDTO();
                eventCustomDTO.getId() = eventDtoClass.getId();
                eventCustomDTOList.add(eventCustomDTO);
            }
            eventCustomDTO.getEventType().add(new TypesOfEvents(eventDtoClass.getEvenTypeId(),eventDtoClass.getColor()));
        }
        return eventCustomDTOList;
    }

Answer 2

由于您的预期结果中有不同的 JSON，因此您需要有 2 个类来序列化/反序列化。让我们考虑以下类：

@Getter
@AllArgsConstructor
private class ObjectA {
    public Long id;
    public String color;
}

@ToString
@AllArgsConstructor
private class ObjectB {
    public Long id;
    public String eventType;
}

现在在这种情况下获得预期结果的代码将是这样的：

List<ObjectA> object = List.of(new ObjectA(1L, "green"), new ObjectA(1L, "red"), new ObjectA(2L, "yellow"));
object.stream().collect(Collectors.groupingBy(ObjectA::getId))
        .entrySet() // from map of grouped ObjectA's
        .stream()
        .map(entry -> new ObjectB(entry.getKey(), // GroupID
                                  entry.getValue() // list of grouped ObjectA's by current GroupID
                                         .stream()
                                         .map(ObjectA::getColor) // change to stream of all color strings from grouped ObjectA's by current GroupID
                                         .collect(Collectors.joining(", ")))
        ).forEach(System.out::println);
        //.collect(Collectors.toList());

有输出：

ObjectB(id=1, eventType=green, red)
ObjectB(id=2, eventType=yellow)

您可以根据您的用例简单地调整该代码。