我使用Decodable协议解码一些json,但我遇到了一个问题:
我得到了一个回答,如果没有添加到元素的地理位置数据,则经度和纬度可以是整数(纬度= 0),也可以是字符串(fx。纬度=" 25.047880")是否有可用的地理数据。现在,当我解码json时,我不知道如何构建我的Struct,因为long和lat都不能是String和Int ..所以我在获取元素时遇到解码错误其中两个案例都有代表。
有关如何解决此问题的任何建议?我已尝试过" Any"作为数据类型,但这不符合可解码协议
struct JPhoto: Decodable {
let id: String
let farm: Int
let secret: String
let server: String
let owner: String
let title: String
let latitude: String //Can both be Int and String
let longitude: String //Can both be Int and String
}
答案 0 :(得分:0)
您需要编写自己的编码器/解码器。您可以使用关联的值枚举来执行此操作,使用switch语句进行编码以及要解码的throw / catch行为:
enum AngularDistance:Codable {
case string(String), integer(Int)
func encode(to encoder: Encoder) throws {
switch self {
case .string(let str):
var container = encoder.singleValueContainer()
try container.encode(str)
case .integer(let int):
var container = encoder.singleValueContainer()
try container.encode(int)
}
}
init(from decoder: Decoder) throws {
do {
let container = try decoder.singleValueContainer()
let str = try container.decode(String.self)
self = AngularDistance.string(str)
}
catch {
do { let container = try decoder.singleValueContainer()
let int = try container.decode(Int.self)
self = AngularDistance.integer(int)
}
catch {
throw DecodingError.typeMismatch(AngularDistance.self, DecodingError.Context(codingPath: decoder.codingPath, debugDescription: "Expected to decode an Int or a String"))
}
}
}
}
以下是对此AngularDistance
类型进行编码和解码的示例:
let lat = [AngularDistance.string("String"), AngularDistance.integer(10)]
let encoder = JSONEncoder()
var decoder = JSONDecoder()
do {
let encoded = try encoder.encode(lat)
try decoder.decode(Array<AngularDistance>.self, from: encoded)
}
catch DecodingError.typeMismatch(let t, let e) {
t
e.codingPath
e.debugDescription
}
catch {
print(error.localizedDescription)
}
这是你的结构重写:
struct JPhoto: Decodable {
let id: String
let farm: Int
let secret: String
let server: String
let owner: String
let title: String
let latitude: AngularDistance //Can both be Int and String
let longitude: AngularDistance //Can both be Int and String
}