如何用ArduinoJson 6创建含4个对象的JSON数组？无需降级

我尝试用ArduinoJson 6生成一个包含4个同结构对象的JSON数据，但一直失败。ArduinoJson 5的.createObject方法用起来更顺手，想知道v6为啥移除了这个方法，有没有替代实现，是不是必须降级到v5？

预期生成的JSON格式 #

{
   "Spot":[
      {
         "LocationId":"673d855c-9f66-4e49-8b2c-737e829d880c",
         "SpotId":"1",
         "Occupied":"Empty"
      },
      {
         "LocationId":"673d855c-9f66-4e49-8b2c-737e829d880c",
         "SpotId":"2",
         "Occupied":"Empty"
      },
      {
         "LocationId":"673d855c-9f66-4e49-8b2c-737e829d880c",
         "SpotId":"3",
         "Occupied":"Empty"
      },
      {
         "LocationId":"673d855c-9f66-4e49-8b2c-737e829d880c",
         "SpotId":"4",
         "Occupied":"Empty"
      }
   ]
}

当前代码 #

char json_string[256];
StaticJsonDocument<256> carparkResponse;

JsonObject spot1 = carparkResponse.createNestedObject("Spot");
spot1["LocationId"] = LOCATION_ID;
spot1["SpotId"] = "1";
spot1["Occupied"] = occupied1.isOccupied();

JsonObject spot2 = spot1.createNestedObject();
spot2["LocationId"] = LOCATION_ID;
spot2["SpotId"] = "2";
spot2["Occupied"] = occupied2.isOccupied();

JsonObject spot3 = spot2.createNestedObject();
spot3["LocationId"] = LOCATION_ID;
spot3["SpotId"] = "3";
spot3["Occupied"] = occupied3.isOccupied();

JsonObject spot4 = spot3.createNestedObject();
spot4["LocationId"] = LOCATION_ID;
spot4["SpotId"] = "4";
spot4["Occupied"] = occupied4.isOccupied();

serializeJson(carparkResponse, json_string);
Serial.println(json_string);

1. 当前代码的核心问题 #

你错误地将Spot创建成了嵌套对象，而非目标格式的JSON数组。createNestedObject("Spot")会在根文档生成一个名为Spot的对象，后续调用createNestedObject()又会在这个对象里嵌套子对象，最终得到的是多层嵌套结构，完全不符合预期的数组格式。

2. ArduinoJson 6 移除.createObject的原因 #

v6移除该方法是为了简化API、降低内存开销并减少逻辑错误。v5的.createObject需要手动管理对象归属，容易出现内存泄漏或结构混乱；v6通过JsonDocument统一管理所有节点内存，改用createNestedObject()（创建嵌套对象）和createNestedArray()（创建嵌套数组）明确区分结构类型，逻辑更清晰，性能更优。

3. ArduinoJson 6 的正确实现方式 #

要生成预期格式，需先创建Spot数组，再往数组中逐个添加对象：

改进后的代码 #

char json_string[256];
StaticJsonDocument<256> carparkResponse;

// 创建Spot数组
JsonArray spotArray = carparkResponse.createNestedArray("Spot");

// 添加第一个车位对象
JsonObject spot1 = spotArray.createNestedObject();
spot1["LocationId"] = LOCATION_ID;
spot1["SpotId"] = "1";
spot1["Occupied"] = occupied1.isOccupied() ? "Occupied" : "Empty";

// 添加第二个车位对象
JsonObject spot2 = spotArray.createNestedObject();
spot2["LocationId"] = LOCATION_ID;
spot2["SpotId"] = "2";
spot2["Occupied"] = occupied2.isOccupied() ? "Occupied" : "Empty";

// 添加第三个车位对象
JsonObject spot3 = spotArray.createNestedObject();
spot3["LocationId"] = LOCATION_ID;
spot3["SpotId"] = "3";
spot3["Occupied"] = occupied3.isOccupied() ? "Occupied" : "Empty";

// 添加第四个车位对象
JsonObject spot4 = spotArray.createNestedObject();
spot4["LocationId"] = LOCATION_ID;
spot4["SpotId"] = "4";
spot4["Occupied"] = occupied4.isOccupied() ? "Occupied" : "Empty";

serializeJson(carparkResponse, json_string);
Serial.println(json_string);

简化循环写法 #

如果车位数量固定，可通过循环减少重复代码：

char json_string[256];
StaticJsonDocument<256> carparkResponse;
JsonArray spotArray = carparkResponse.createNestedArray("Spot");

// 存储所有车位的占用状态
bool occupiedStates[] = {occupied1.isOccupied(), occupied2.isOccupied(), occupied3.isOccupied(), occupied4.isOccupied()};

for (int i = 0; i < 4; i++) {
  JsonObject spot = spotArray.createNestedObject();
  spot["LocationId"] = LOCATION_ID;
  spot["SpotId"] = String(i + 1);
  spot["Occupied"] = occupiedStates[i] ? "Occupied" : "Empty";
}

serializeJson(carparkResponse, json_string);
Serial.println(json_string);

4. 是否需要降级到v5？ #

完全不需要。ArduinoJson 6的API设计更合理，内存管理和性能都优于v5，只要理清数组与对象的创建逻辑，就能轻松实现需求。

内容的提问来源于stack exchange，提问作者Corey673