快速将 Codable/Encodable 转换为 JSON 对象

Question

最近我加入Codable了一个项目并JSON从符合Encodable我的类型中获取一个对象，我想出了这个扩展，

extension Encodable {

    /// Converting object to postable JSON
    func toJSON(_ encoder: JSONEncoder = JSONEncoder()) -> [String: Any] {
        guard let data = try? encoder.encode(self),
              let object = try? JSONSerialization.jsonObject(with: data, options: .allowFragments),
              let json = object as? [String: Any] else { return [:] }
        return json
    }
}

这很有效，但有没有更好的方法来实现同样的目标？

Answer 1

使用此扩展将可编码对象转换为 JSON 字符串：

extension Encodable {
    /// Converting object to postable JSON
    func toJSON(_ encoder: JSONEncoder = JSONEncoder()) throws -> NSString {
        let data = try encoder.encode(self)
        let result = String(decoding: data, as: UTF8.self)
        return NSString(string: result)
    }
}

Answer 2

我的建议是命名函数toDictionary并将可能的错误交给调用者。有条件的低垂失败（类型不匹配），则抛出包裹在typeMismatch 德编码误差。

extension Encodable {

    /// Converting object to postable dictionary
    func toDictionary(_ encoder: JSONEncoder = JSONEncoder()) throws -> [String: Any] {
        let data = try encoder.encode(self)
        let object = try JSONSerialization.jsonObject(with: data)
        guard let json = object as? [String: Any] else {
            let context = DecodingError.Context(codingPath: [], debugDescription: "Deserialized object is not a dictionary")
            throw DecodingError.typeMismatch(type(of: object), context)
        }
        return json
    }
}