无法分配“ NSDictionary”类型的值？ 键入“字符串？”

Question

下面是我从API获取响应的代码（Swift 4.2.1），但是在调用时显示了一些错误，

无法分配“ NSDictionary”类型的值？ 键入“字符串？”

let task = URLSession.shared.dataTask(with: request) { data, response, error in
   guard let data = data, error == nil else {
      return
   }
   let responseString = String(data: data, encoding: .utf8)
   print("responseString = \(String(describing: responseString))")
   do {
    let jsonResponse = try JSONSerialization.jsonObject(with: data, options: []) as? NSDictionary
     print(jsonResponse!)
     responsevalue = jsonResponse

     let alert1 = UIAlertController(title: "Alert", message: self.responsevalue, preferredStyle: UIAlertController.Style.alert)
     let action1 = UIAlertAction(title: "Ok",style: .default)
     {(ACTION)in
     }
     alert1.addAction(action1)
     self.present(alert1, animated: true, completion: nil)
     }catch let parsingError {
        print("Error", parsingError)

        let alert1 = UIAlertController(title: "Alert", message: parsingError as? String, preferredStyle: UIAlertController.Style.alert)
        let action1 = UIAlertAction(title: "Ok",style: .default)
         {(ACTION)in
         }
         alert1.addAction(action1)
         self.present(alert1, animated: true, completion: nil)
         }
       }
      task.resume()

Answer 1

不要使用NSDictionary 。 使用[String: Any] 。 您需要从字典中获取消息字符串以显示在警报中。

let jsonResponse = try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
print(jsonResponse)
responsevalue = jsonResponse
guard let message = jsonResponse["msg"] as? String else {//Use the key from json
    return
}
let alert1 = UIAlertController(title: "Alert", message: message, preferredStyle: .alert)

在其他部分，请使用parsingError.localizedDescription而不是parsingError

let alert1 = UIAlertController(title: "Alert", message: parsingError.localizedDescription, preferredStyle: .alert)

Answer 2

如错误所示， Cannot assign value of type 'NSDictionary?' to type 'String?' Cannot assign value of type 'NSDictionary?' to type 'String?' 。 您正在将NSDictionary分配给类型为String的变量。

您需要将responseString的原始声明更改为NSDictionary或另一个变量以存储NSDictionary。