Swift 解码具有多种不同格式的数据类型

Question

我从服务器返回一个 "several" 不同格式的布尔值（对于相同的结构和字段）。我知道这很荒谬，但我需要找到一种方法来干净地处理它。

所以为了反序列化它，我做了类似的事情（示例程序）：

import Foundation

struct Foo: Codable {
    var isOpen: Bool?

    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        isOpen = try container.decodeIfPresent(Bool.self, forKey: .isOpen)
    }

    enum CodingKeys: String, CodingKey {
        case isOpen
    }
}

//He sends any one of these..
let json1 = "{ \"isOpen\": \"true\" }"
let json2 = "{ \"isOpen\": \"false\" }"
let json3 = "{ \"isOpen\": true }"
let json4 = "{ \"isOpen\": false }"
let json5 = "{ \"isOpen\": null }"
let json6 = "{ \"isOpen\": \"null\" }"
let json7 = "{ \"isOpen\": \"<null>\" }"

//He doesn't send this one.. but I wouldn't be surprised if I got it so I added it for fun (serializing the below `json8` and `json9` is not required for an answer).. :)

let json8 = "{ \"isOpen\": 0 }"
let json9 = "{ \"isOpen\": 1 }"

let json = [json1, json2, json3, json4, json5, json6, json7, json8, json9]
for js in json {
    if let rawData = js.data(using: .utf8) {
        do {
            let foo = try JSONDecoder().decode(Foo.self, from: rawData)
            if let isOpen = foo.isOpen {
                print("\(isOpen)\n\n")
            } else {
                print("State Unknown\n\n")
            }
        } catch {
            print("\(error)\n\n")
        }
    }
}

现在，如果我使用 Swift Codable（我们所有的数据结构都已使用），那么我们将得到类型不匹配并抛出 error/exception。我考虑过尝试捕获每个案例并尝试使用不同类型的另一个解码但最终会像：

do {
    isOpen = try container.decode(Bool.self, forKey: .isOpen)
}
catch {
    do {
        isOpen = try container.decode(Int.self, forKey: .isOpen) != 0
    }
    catch {
        do {
            isOpen = Bool(try container.decode(String.self, forKey: .isOpen))
        }
        catch {
            do {
                isOpen = try container.decodeIfPreset(Bool.self, forKey: .isOpen)  ?? GiveUpAndAssignDefaultValueHere..
            }
            catch {
                isOpen = nil //no idea..
            }
        }
    }
}

然后这让我开始考虑先将它转换为字符串，然后尝试解析它，所以我最终得到了（至少比上面的更好）：

do {
    isOpen = try container.decode(Bool?.self, forKey: .isOpen)
}
catch {
    do {
        isOpen = Bool(try container.decode(String.self, forKey: .isOpen))
    }
    catch {
        isOpen = try container.decode(Int.self, forKey: .isOpen) != 0
    }
}

但肯定有更好的方法吗？有什么想法吗？？？

Answer 1

一个新建议是使用相同的 CodingKey 解码多值 isOpen

代码应该是这样的，

struct Foo: Codable {
var isOpen: Bool?
private var isOpenInty: Int?
private var isOpenStringy: String?
init(from decoder: Decoder) throws {
    let container = try decoder.container(keyedBy: CodingKeys.self)
    do {
        isOpen = try container.decodeIfPresent(Bool.self, forKey: .isOpen)

    }catch {
        do {
            isOpenInty = try container.decodeIfPresent(Int.self, forKey: .isOpen)
            if isOpenInty == 0  {isOpen = true} else {isOpen = false}
        }catch {
            isOpenStringy = try container.decodeIfPresent(String.self, forKey: .isOpen)
            if isOpenStringy == "true" {isOpen = true} else {isOpen = false}
        }
    }
}

并根据这些值中的任何一个，设置 isOpen 值。

只是处理这种情况的另一种方式。

这里和你的想法差不多，

   do {
        isOpen = try container.decode(Bool?.self, forKey: .isOpen)
    }
    catch {
        do {
            isOpen = Bool(try container.decode(String.self, forKey: .isOpen))
        }
        catch {
            isOpen = try container.decode(Int.self, forKey: .isOpen) != 0
        }
    }

但是你的代码在 null 的情况下给出 State Unknown，如果 null，它上面的代码简单地处理它，作为 false

Answer 2

而不是catch错误我有条件地绑定类型

struct Foo: Codable {
    var isOpen: Bool?

    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        if let boolOpen = try? container.decode(Bool.self, forKey: .isOpen) {
            isOpen = boolOpen
        } else if let intOpen = try? container.decode(Int.self, forKey: .isOpen) {
            isOpen = intOpen == 1
        } else if let stringOpen = try? container.decode(String.self, forKey: .isOpen) {
            switch stringOpen {
            case "true", "1": isOpen = true
            case "false", "0": isOpen = false
            default : isOpen = nil
            }
        } else {
            isOpen = nil
        }
    }
}

Answer 3

另一种方法，原理相同，仅供娱乐。非常O.K.

    init(from decoder: Decoder) throws {

let container = try decoder.container(keyedBy: CodingKeys.self)
_  =  [ try? container.decode(Bool.self, forKey: .isOpen),
    try?  container.decode(String.self, forKey: .isOpen),
    try? container.decode(Int.self, forKey: .isOpen)].first{
switch [=10=] {
case is Bool:
    self.isOpen  = [=10=] as? Bool
    return true
case is Int:
    self.isOpen  = ([=10=] as! Int) == 0 ? false : (([=10=] as! Int) == 1 ? true : nil)
    return true
case is String:
    self.isOpen  = Bool.init([=10=] as! String)
    return true
default:
    return false
}
    }
}

Swift 解码具有多种不同格式的数据类型

Swift Decode a data type with many different formats

ios

codable

decodable

swift4.2