假设您有一些 JSON:
{
"status": "error",
"data": {
"errormessage": "Could not get user with ID: -1.",
"errorcode": 14
}
}
对于给定的错误结构:
struct APIError: Decodable {
let code: Int?
let message: String?
enum CodingKeys: String, CodingKey {
case code = "errorcode"
case message = "errormessage"
}
}
点击 Web 服务,获取 JSON,然后初始化结构:
let urlRequest = URLRequest(url: url)
let session = URLSession.shared
let task = session.dataTask(with: urlRequest)
{ (data, response, error) in
// Doesn't work because the portion of the JSON we want is in the "data" key
let e = try? JSONDecoder().decode(APIError.self, from: data)
}
task.resume()
有没有一些简单的方法可以做类似的事情data["data"]
?要遵循的正确模型是什么?
解决方案 A - 将数据转换为 JSON 对象,获取我们想要的对象,然后将其转换为 Data 对象并进行解码。
let jsonFull = (try? JSONSerialization.jsonObject(with: data, options: [])) as? [String: Any]
let json = jsonFull["data"]
let data_error = try? JSONSerialization.data(withJSONObject: json, options: [])
let e = try? JSONDecoder().decode(APIError.self, from: data_error)
解决方案 B - 将目标项包装在另一个结构中
struct temp : Decodable {
let status: String?
let data: APIError?
}
let e = try? JSONDecoder().decode(temp.self, from: data).data
解决方案 C - 在 decode 中设置嵌套结构(如果它是几个对象深怎么办?)
let e = try? JSONDecoder().decode([Any, APIError.self], from: data)
我错过了什么模式?最优雅的方法是什么?