无法分配“ NSDictionary”类型的值?键入“字符串?”

时间:2019-05-20 11:16:13

标签: ios arrays json swift

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

  

无法分配类型为“ NSDictionary?”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()

2 个答案:

答案 0 :(得分:3)

请勿使用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)

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

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

答案 1 :(得分:0)

,因为错误显示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。