解码 Swift 中包含特殊字符的网址

Question

我使用的 API 提供 URL 链接，可以包含特殊字符，如“http://es.dbpedia.org/resource/Análisis_de_datos ”（里面的字母“á”）。

这是一个绝对有效的 URL，但是，如果可解码的 class 包含可选的 URL? 变量，则无法解码。

我可以在我的 class 中将 URL? 更改为 String? 并使用类似 URL(string: urlString.addingPercentEncoding(withAllowedCharacters: .urlFragmentAllowed) 的计算变量，但也许有更优雅的解决方案。

要在 Playground 中重现：

struct Container: Encodable {
    let url: String
}

struct Response: Decodable {
    let url: URL?
}

let container = Container(url: "http://es.dbpedia.org/resource/Análisis_de_datos")

let encoder = JSONEncoder()
let encodedData = try encoder.encode(container)
    
let decoder = JSONDecoder()
let response = try? decoder.decode(Response.self, from: encodedData)
// response == nil, as it can't be decoded.

let url = response?.url

Answer 1

您可以扩展 KeyedDecodingContainer 并实现您自己的 URL 解码方法：

extension KeyedDecodingContainer {
    func decode(_ type: URL.Type, forKey key: K) throws -> URL {
        let string = try decode(String.self, forKey: key)
        guard let url = URL(string: string.addingPercentEncoding(withAllowedCharacters: .urlFragmentAllowed) ?? "")
        else {
            throw DecodingError.dataCorrupted(.init(codingPath: codingPath, debugDescription: "The stringvalue for the key \(key) couldn't be converted into a URL value: \(string)"))
        }
        return url
    }
    // decoding an optional URL
    func decodeIfPresent(_ type: URL.Type, forKey key: K) throws -> URL? {
        try URL(string: decode(String.self, forKey: key).addingPercentEncoding(withAllowedCharacters: .urlFragmentAllowed) ?? "")
    }
}

struct Container: Encodable {
    let url: String
}

struct Response: Decodable {
    let url: URL
}

let container = Container(url: "http://es.dbpedia.org/resource/Análisis_de_datos")

do {
    let encodedData = try encoder.encode(container)
    print(String(data: encodedData, encoding: .utf8))
    let decoder = JSONDecoder()
    let response = try decoder.decode(Response.self, from: encodedData)
    print(response)
} catch {
    print(error)
}

这将打印：

Optional("{"url":"http:\/\/es.dbpedia.org\/resource\/Análisis_de_datos"}")

Response(url: http://es.dbpedia.org/resource/An%C3%A1lisis_de_datos)

Answer 2

有多种方法可以解决这个问题，但我认为使用属性包装器可能是最优雅的：

@propertyWrapper
struct URLPercentEncoding {
   var wrappedValue: URL
}

extension URLPercentEncoding: Decodable {
   public init(from decoder: Decoder) throws {
      let container = try decoder.singleValueContainer()
        
      if let str = try? container.decode(String.self),
         let encoded = str.addingPercentEncoding(
                              withAllowedCharacters: .urlFragmentAllowed),
         let url = URL(string: encoded) {

         self.wrappedValue = url

      } else {
         throw DecodingError.dataCorrupted(
            .init(codingPath: container.codingPath, debugDescription: "Corrupted url"))
      }
   }
}

然后你就可以像这样使用它，而这个模型的消费者不必知道任何关于它的信息：

struct Response: Decodable {
    @URLPercentEncoding let url: URL
}

解码 Swift 中包含特殊字符的网址

Decode URLs with special characters in Swift

url

json

swift

codable

decodable