Question

我需要使用嵌套对象创建答案JSON

我尝试使用mappable，但无法解决问题

我需要创建这种数组

{  
   "interactionId":"daade6b6adcd8b063a355e28bc1f1341",
   "futureSurveyAnswers":[  
      {  
         "type":"imagepicker",
         "qcode":"vxo20zeezo",
         "values":[  
            {  
               "value":"lion"
            }
         ]
      },
      {  
         "type":"radiogroup",
         "qcode":"s4ep4s0shf",
         "values":[  
            {  
               "value":"item1"
            }
         ]
      },
      {  
         "type":"checkbox",
         "qcode":"76k5cpnpki",
         "values":[  
            {  
               "value":"item1"
            },
            {  
               "value":"item2"
            }
         ]
      }
   ],
   "originalResultArray":"{\"question2\":\"item1\",\"question3\":[\"item1\",\"item2\"],\"question1\":\"lion\"}"
}

Answer 1

当前最好使用的最佳解决方案是Codable协议。在线上有很多示例和教程，因此请检查一两个，我相信您会发现它非常简单。

关键是您可以直接在JSON中编码具体的类或结构。因此，您的问题就不在JSON本身，而只是结构化您的类。例如：

class Survey {

    class Answer {
        enum AnswerType: String {
            case imagepicker
        }
        class Value {
            var value: String?
        }

        var type: AnswerType = .imagepicker
        var qcode: String?
        var values: [Value] = [Value]()
    }

    let interactionId: String
    var futureSurveyAnswers: [Answer] = [Answer]()
    var originalResultArray: String?

    init(interactionId: String) { self.interactionId = interactionId }
}

这样，您可以轻松地根据需要修改结构。例如：

let survey = Survey(interactionId: "0")
survey.futureSurveyAnswers.append({
    let answer = Survey.Answer()
    answer.qcode = "test"
    return answer
}())

因此，要将其扩展到可编码，您实际上需要将Codable协议附加到您的类和任何嵌入式组件上：

class Survey: Codable {

    class Answer: Codable {
        enum AnswerType: String, Codable {
            case imagepicker
        }
        class Value: Codable {
            var value: String?
        }

        var type: AnswerType = .imagepicker
        var qcode: String?
        var values: [Value] = [Value]()
    }

    let interactionId: String
    var futureSurveyAnswers: [Answer] = [Answer]()
    var originalResultArray: String?

    init(interactionId: String) { self.interactionId = interactionId }
}

现在您可以像以下这样简单地获取JSON：

let survey = Survey(interactionId: "0")
survey.futureSurveyAnswers.append({
    let answer = Survey.Answer()
    answer.values.append({
        let value = Survey.Answer.Value()
        value.value = "some"
        return value
    }())
    answer.qcode = "test"
    return answer
}())

let encoder = JSONEncoder()
if let jsonData = try? encoder.encode(survey) {
    print("Generated data: \(jsonData)")
}

我希望这能为您指明正确的方向。

在评论中，这就是我用作JSON的测试示例：

class Survey: Codable {

    class Answer: Codable {
        enum AnswerType: String, Codable {
            case imagepicker
        }
        class Value: Codable {
            var value: String?
            init(value: String? = nil) { self.value = value }
        }

        var type: AnswerType = .imagepicker
        var qcode: String?
        var values: [Value] = [Value]()
    }

    let interactionId: String
    var futureSurveyAnswers: [Answer] = [Answer]()
    var originalResultArray: String?

    init(interactionId: String) { self.interactionId = interactionId }
}

func addTestAnswer(to survey: Survey) {
    let answer = Survey.Answer()
    answer.values.append(.init(value: "Random value \(Int.random(in: 1...100))"))
    answer.values.append(.init(value: "Random value \(Int.random(in: 1...100))"))
    answer.values.append(.init(value: "Random value \(Int.random(in: 1...100))"))
    answer.values.append(.init(value: "Random value \(Int.random(in: 1...100))"))
    survey.futureSurveyAnswers.append(answer)
}

func testRandomSurveyJSON() {
    let survey = Survey(interactionId: "randomSurvey")
    addTestAnswer(to: survey)
    addTestAnswer(to: survey)
    addTestAnswer(to: survey)
    addTestAnswer(to: survey)
    addTestAnswer(to: survey)

    print("Got JSON: \(String(data: try! JSONEncoder().encode(survey), encoding: .utf8)!)")
}

并得到以下结果：

Got JSON: {"interactionId":"randomSurvey","futureSurveyAnswers":[{"type":"imagepicker","values":[{"value":"Random value 74"},{"value":"Random value 4"},{"value":"Random value 26"},{"value":"Random value 93"}]},{"type":"imagepicker","values":[{"value":"Random value 43"},{"value":"Random value 65"},{"value":"Random value 38"},{"value":"Random value 88"}]},{"type":"imagepicker","values":[{"value":"Random value 56"},{"value":"Random value 88"},{"value":"Random value 57"},{"value":"Random value 94"}]},{"type":"imagepicker","values":[{"value":"Random value 66"},{"value":"Random value 52"},{"value":"Random value 89"},{"value":"Random value 27"}]},{"type":"imagepicker","values":[{"value":"Random value 53"},{"value":"Random value 93"},{"value":"Random value 30"},{"value":"Random value 55"}]}]}

这似乎是正确的。

如何快速将json对象逐个写入json对象

1 个答案: