Swift //将String Json文件转换为枚举大小写

Question

我有以下对象:

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)

如何将Json String"dishType"转换为我用enum创建的"DishType"类型,以便正确填充我的Food实例？

Answer 1

您可能希望为枚举指定关联值:

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

或者,更简单:

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

然后你可以这样做:

dishType = DishType(rawValue: string)

例如

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

就个人而言,如果做Swift 4,我会退休SwiftyJSON并使用本机JSONDecoder并声明你的类型Codable.(注意,我们仍然需要定义DishType具有相关值,如上所述.)

例如,让我们假设您的回答是这样的:

{
    "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"
        }
    ]
}

然后你可以像这样定义你的类型:

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
}

然后你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)
        }
}

这使您完全脱离了手动迭代结果以将其映射到对象的业务.

显然,我认为你的真实响应有更多的键而不仅仅是foods,所以你要添加你需要的任何字段FoodsResponse,但希望这说明了让JSONDecoderJSON自动解析到模型结构的想法.

有关JSONDecoder和Codable类型的更多信息,请参阅编码和解码自定义类型.

顺便说一句,我的示例FoodResponse结构提示了一些问题,为什么我不只是假设Web服务将返回一个Food对象数组.让我解释一下我的理由.

FoodsResponseWeb服务响应中更典型的结构类似于:

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"
}

我认为最好有一个包含一些常见成功布尔的结构,例如success,所有格式良好的响应包括,然后分别填充成功或失败的各种属性.