json - 将 DataResponse<String> 转换为可用类型

Question

我有一个 POST 请求，如下所示。我需要一些有关如何使用响应数据的帮助。我得到的答复是字典的形式。这里使用的最佳实践是什么？我收到的错误消息是:

Cannot assign value of type 'DataResponse' to type 'String'

func getRequest(){
    let urlString = "http://scissors.pythonanywhere.com/getRequest"
    Alamofire.request(urlString, method: .post, parameters: ["date": self.weekDays[self.itemSelectedIndex]],encoding: JSONEncoding.default, headers: nil).responseString {
        response in
        switch response.result {
        case .success:
            print(response)
            var backToString = String(data: response, encoding: String.Encoding.utf8) as String?

            break
        case .failure(let error):

            print(error)
        }
    }
}

我尝试了 self.bookings = 响应为! String 但它给了我一个警告，说它会失败。也尝试过这个:Alamofire in Swift: converting response data in usable JSON dictionary运气不好。

编辑

这是我从响应中得到的打印结果。

SUCCESS: {
    "10:00" =     {
        booked = false;
        name = "";
        number = "";
    };
    "10:30" =     {
        booked = false;
        name = "";
        number = "";
    };
    "11:00" =     {
        booked = false;
        name = "";
        number = "";
    };
}

Answer 1

我实际上最终使用了这个非常简单易行的方法。就我而言，预订的声明如下: var bookings: NSDictionary = [:]

func getRequest(){

    let urlString = "http://scissors.pythonanywhere.com/getRequest"
    Alamofire.request(urlString, method: .post, parameters: ["date": self.weekDays[self.itemSelectedIndex]],encoding: JSONEncoding.default, headers: nil).responseJSON {
        response in
        switch response.result {
        case .success(let JSON):
            self.bookings = JSON as! NSDictionary

            self.tableView.reloadData()

            break
        case .failure(let error):
            print(error)
        }
    }
}