从 JSON 中解析十进制显示为字符串

Question

使用 Xcode 10.2 和 iOS 12.x 我们能够从 json 字符串中提取小数。 Xcode 11.1 和 iOS 13.1 抛出异常

Expected to decode Double but found a string/data instead.

class MyClass : Codable {

     var decimal: Decimal?
 }

然后尝试解析它

let json = "{\"decimal\":\"0.007\"}"
let data = json.data(using: .utf8)
let decoder = JSONDecoder()
decoder.nonConformingFloatDecodingStrategy = .convertFromString(positiveInfinity: "s1", negativeInfinity: "s2", nan: "s3")
 do {
   let t = try decoder.decode(MyClass.self, from: data!)
 } catch {
   print(error)
 }

如果我将 json 字符串更改为

let json = "{\"decimal\":0.007}"

它有效，但我们又一次失去了精度。有什么想法吗？

Answer 1

我认为更简洁的解决方案是声明值，而不是像字符串那样声明值：

"test": 0.007

具有这样的结构：

struct Stuff {
     var test: Decimal
}

然后：

let decoder = JSONDecoder()
let stuff = try decoder.decode(Stuff.self, from: json)

否则你可以使用这个例子：

https://forums.swift.org/t/parsing-decimal-values-from-json/6906/3

Answer 2

该解码策略与将数字表示为字符串无关。你需要做的是实现 init(from:) 并从那里的字符串转换

class MyClass : Codable {
    var decimal: Double?

    enum CodingKeys: String, CodingKey {
        case decimal = "test"
    }

    required init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        decimal = Double(try container.decode(String.self, forKey: .decimal)
        //or if Decimal is used:
        //decimal = Decimal(string: try container.decode(String.self, forKey: .decimal)
    }
}

请注意，我在这里使用 Double 而不是 Decimal 以使其更简单

Answer 3

struct Root: Codable {
    let decimal: Decimal
}

extension Root {
    public init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        decimal = try Decimal(string: container.decode(String.self, forKey: .decimal)) ?? .zero
    }
}

let json = #"{"decimal":"0.007"}"# 
do {
    let root = try JSONDecoder().decode(Root.self, from: .init(json.utf8))
    print(root)
} catch {
    print(error)
}

这将打印

Root(decimal: 0.007)

Answer 4

类型应该是 Double 并且在解析中也定义为 Double。 Swift 会弄清楚剩下的

struct MyClass: Decodable {
        let decimal: Double

        //can be renamed to follow the API name.
        enum CodingKeys: String, CodingKey {
            case decimal
        }
    }
    extension MyClass {
        init(from decoder: Decoder) throws {

            let values = try decoder.container(keyedBy: CodingKeys.self)
            decimal = try values.decode(Double.self, forKey: .decimal)

        }
    }

Answer 5

您需要扩展 KeyedDecodingContainer 并为 Decimal.Type 添加一个实现。

extension KeyedDecodingContainer {
    func decode(_ type: Decimal.Type, forKey key: K) throws -> Decimal {
        let stringValue = try decode(String.self, forKey: key)
        guard let decimalValue = Decimal(string: stringValue) else {
            let context = DecodingError.Context(codingPath: [key], debugDescription: "The key \(key) couldn't be converted to a Decimal value")
            throw DecodingError.typeMismatch(type, context)
        }
        return decimalValue
    }
}

这是一个例子：

let json = """
{
  "capAmount": "123.45"
}
"""

struct Status: Decodable {
    let capAmount: Decimal

    enum CodingKeys: String, CodingKey {
        case capAmount
    }

    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        capAmount = try container.decode(Decimal.self, forKey: .capAmount)
    }
}

// Execute it
if let data = json.data(using: .utf8){
    let status = try JSONDecoder().decode(Status.self, from: data)
    print(status.capAmount)
}

从 JSON 中解析十进制显示为字符串

Parsing Decimal from JSON presented as string

swift

codable

ios13