我有下面的对象:

class Food {
    var cal: Int
    var displayName: String
    var imgUrl: String
    var dishType: DishType

    init(cal: Int, displayName: String, imgUrl: String, dishType: DishType) {
        self.cal = cal
        self.displayName = displayName
        self.imgUrl = imgUrl
        self.dishType = dishtype
    }
}

enum DishType {
    case starter
    case main
    case desert
}

这是我的Alamofire请求的一部分:
if let cal = foodJson["cal"].int,
    let displayName = foodJson["display_name"].string,
    let dishType = foodJson["type"].string,
    let imgUrl = foodJson["imgUrl"].string {
    let food = Food(cal: cal, displayName: displayName, imgUrl: imgUrl, dishType: ??)

    foods.append(food)

我如何将Json字符串“dishType”转换为用枚举创建的“DishType”类型,以便正确填充我的Food实例?

最佳答案

您可能想为您的枚举指定一个关联值:

enum DishType: String {
    case starter = "starter"
    case main    = "main"
    case desert  = "desert"
}

或者,更简单地说:
enum DishType: String {
    case starter
    case main
    case desert
}

然后,您可以执行以下操作:
dishType = DishType(rawValue: string)

例如
if let dishTypeString = foodJson["type"].string,
    let dishType = DishType(rawValue: dishTypeString) {
        ...
}

就个人而言,如果使用Swift 4,我将退休SwiftyJSON并使用本机JSONDecoder并将您的类型声明为Codable。 (请注意,我们仍然需要定义DishType以具有关联的值,如上所述)。

例如,假设您的响应是这样的:
{
    "foods": [{
            "cal": 800,
            "display_name": "Beef",
            "imgUrl": "http://example.com/wheres_the_beef.jpg",
            "dishType": "main"
        },
        {
            "cal": 2000,
            "display_name": "Chocolate cake",
            "imgUrl": "http://example.com/yummy.jpg",
            "dishType": "desert"
        }
    ]
}

然后,您可以像这样定义类型:
struct Food: Codable {
    let cal: Int
    let displayName: String
    let imgUrl: String
    let dishType: DishType
}

enum DishType: String, Codable {
    case starter
    case main
    case desert
}

然后您可以像这样解析响应:
struct FoodsResponse: Codable {
    let foods: [Food]
}

Alamofire.request(url)
    .responseData { response in
        switch response.result {
        case .success(let data):
            do {
                let decoder = JSONDecoder()
                decoder.keyDecodingStrategy = .convertFromSnakeCase
                let responseObject = try decoder.decode(FoodsResponse.self, from: data)

                print(responseObject.foods)
            } catch {
                print(error)
            }

        case .failure(let error):
            print(error)
        }
}

这使您完全不用手动遍历结果以将其映射到对象。

显然,我假设您的实际响应具有更多的键,而不仅仅是foods,因此您可以将所需的任何字段添加到FoodsResponse,但是希望这可以说明让JSONDecoder自动将JSON解析为模型结构的想法。

有关JSONDecoderCodable类型的更多信息,请参见Encoding and Decoding Custom Types

顺便说一下,我的示例FoodResponse结构提示了一个问题,为什么我不仅仅假定Web服务将返回Food对象的数组。让我解释一下我的理由。

Web服务响应中的FoodsResponse的更典型结构是:
struct FoodsResponse: Codable {
    let success: Bool
    let error: String?   // only supplied if `success` was `false`
    let foods: [Food]?   // only supplied if `success` was `true`
}

在此结构中,此响应对象可以处理成功方案,例如:
{
    "success": true,
    "foods": [...]
}

或失败:
{
    "success": false,
    "error": "No data found"
}

我认为最好是包含一些常见的成功 bool(boolean) 值的结构,例如success,所有格式正确的响应都包含,然后分别填充了用于成功或失败的各种属性。

08-19 11:58