Question

我将Xcode更新为Xcode 9.3，Swift 4.1。试图将模型对象转换为JSON给出了下面的数据

{"Email": "some("xxxxxx")", "Password": "some("xxxxxxx")"}

我正在使用JSONSeriliazer，Serilization代码位于

之下

        if(requestObj != nil){
        let json =  JSONSerializer.toJson(requestObj!)
        request.httpBody = json.data(using: .utf8)
    }

Answer 1

如果您尝试通过投射或String?将Optional<String>（String）转换为非可选String(describing:)，则结果字符串将为Optional("your string here")。如果您需要使其按原样工作，则在将其插入JSON字符串之前，您必须将其转换为非可选值。

您可以考虑使用Codable协议进行JSON操作。在此示例中，您也不确定是否希望电子邮件和密码是可选的。

struct LoginRequest: Codable {
    private enum LoginRequestKey: String, CodingKey {
        case email = "Email"
        case password = "Password"
    }

    var email: String
    var password: String

    init(email: String,
         password: String) {
        self.email = email
        self.password = password
    }

    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: LoginRequestKey.self)
        self.email = try container.decode(String.self, forKey: .email)
        self.password = try container.decode(String.self, forKey: .password)
    }

    func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: LoginRequestKey.self)
        try container.encode(email, forKey: .email)
        try container.encode(password, forKey: .password)
    }

}

let encoder = JSONEncoder()
let decoder = JSONDecoder()

let request = LoginRequest(email: "xxxxxx", password: "yyyyyy")
let json = try! encoder.encode(request)
let object = try! decoder.decode(LoginRequest.self, from: json)

如果需要可选，只需使用try?

JSON中的“some（”value“）”

1 个答案: