Alamofire：如何处理成功响应中的错误

Question

你好，我对Alamofire和json还不太熟悉。我想问一下如何处理成功响应的错误。以下是使用POST请求API的响应：

{
data =     {
    "cabinet_id" = 3289146811;
    "created_at" = "2018-04-11T04:53:43.288023001Z";
    "user_id" = 123abc;
};
}

但有时由于服务器错误，我会收到以下响应。

{
error =     {
    "http_status_code" = 400;
    message = "Unable to proceed";
    type = "rent_request_failed";
};
}

这是我所做的：

let url = "myurl"
let parameters = ["user_id": ""\(self.idPassed)"",
                          "cabinet_id": "\(self.stringPassed)"]

        Alamofire.request(url, method: .post, parameters: parameters, encoding: URLEncoding.default).responseJSON { response in

                if (response.result.isSuccess){

                    if let value = response.result.value {
//now success or failed always navigate to here                            
Helper.helper.switchToNavigationViewController(Navigation: "hundredpercent")
                        print(value)

                    }
                //do your json stuff
                } else if (response.result.isFailure) {
                //Manager your error
                switch (response.error!._code){
                case NSURLErrorTimedOut:
                //Manager your time out error
                break
                case NSURLErrorNotConnectedToInternet:
                //Manager your not connected to internet error
                break
                default:
                    let alertt = UIAlertController(title: "Server Error!", message: "Please try again", preferredStyle: .alert)

                    alertt.addAction(UIAlertAction(title: "OK", style: UIAlertActionStyle.default, handler: {(action:UIAlertAction!) in
//need navigate here if error                        
Helper.helper.switchToNavigationViewController(Navigation: "qrcodescan")

                    }))

                    self.present(alertt, animated: true)
                }
                }
        }

- madhatter989

---

您可以从请求本身验证状态码。而不仅仅是请求，要执行 req.validate(statusCode: 200..<299).responseJSON(completionHandler: { response in。这将接受您的状态码为 200-299 的成功响应，并将其余的视为失败。 - Ankit Jayaswal

谢谢@AnkitJayaswal，它可以工作了。 - madhatter989

Answer 1

在使用Alamofire请求API时，通过添加validate(statusCode: 200..<299)来验证响应状态码。

如果您的HTTP状态码在区间200-299内，则验证成功；否则进入失败情况。

        // Call response handler method of alamofire
        Alamofire.request(url, method: .post, parameters: parameters, encoding: URLEncoding.default).validate(statusCode: 200..<299).responseJSON(completionHandler: { response in

            switch response.result {
            case .success(let data): 
                Logs.DLog(object: "\n Success: \(response)")

                // Do your code here...

            case .failure(let error):
                Logs.DLog(object: "\n Failure: \(error.localizedDescription)")

                // Do your code here...

            }
        })

- Ankit Jayaswal

---

默认情况下，可接受的范围为200...299，因此您不需要指定范围。 - j.krissa

Answer 2

这是我处理响应的一种方式。希望能对您有所帮助。

class func Request(statusCallBack: @escaping (_ statusCode: Int) -> Void, successCallback: @escaping (_ result: AnyObject) -> Void, errorCallback: @escaping (_ error: Error) -> Void) {

Alamofire.request(url, method: method, parameters: params, encoding: encodingType, headers: headers).responseJSON { (responce) in

        // Here checking the status code of response
        if let status = responce.response?.statusCode {
            if status == 401 && AppUserDefault.getAccessToken() != nil {
                User.logout()
                return
            }
            statusCallBack(status)
        }
        switch responce.result {
        case .success(_):
            if let result = responce.result.value {
                successCallback(result as AnyObject)
            } else {
                errorCallback(NSError(domain: "Internal server error", code: 1002, userInfo: nil))
            }
            break
        case .failure(let error) :
            errorCallback(error)
            break
        }
    }
}

- isuru madusanka

---

如果你在alamofire请求中添加validate()来验证API的响应状态码，那么你可以进行如下的//检查响应状态码：if let status = responce.response?.statusCode { if status == 401 && AppUserDefault.getAccessToken() != nil { User.logout() return } statusCallBack(status) }在case .failure(let error) :中。 - j.krissa

Answer 3

使用SwiftyJSON并尝试以下操作

  switch response.result {
  case .success:
    if let value = response.result.value {
      let json = JSON(value)
      if ( response.response!.statusCode == 200 || response.response!.statusCode == 201) {
        let localStatusCode = json["status"].intValue

        if localStatusCode >= 400 {
          if json["errors"] != JSON.null{
            callback(json["errors"].stringValue, nil)
          }else{
              callback(json["message"].stringValue, nil)
          }
        }else {
          callback(nil, json)
        }
      }else if response.response?.statusCode != 200 {
        callback(json["error"].stringValue, nil)
      }
    }
  case .failure(let error):

    NSLog("Failure \(error),\(urlString)")

    if(response.response?.statusCode == 401){

    }
    if  let errorStr = response.data {
      let errorJson = JSON(errorStr)
      if(errorJson != JSON.null){
        if errorJson["error"] != JSON.null{
          callback(errorJson["error"].stringValue, nil)
        }else{
          callback(errorJson["message"].stringValue, nil)
        }
      }else {
        callback(error.localizedDescription,nil)
      }
    }else {
      callback("Error Connecting to Server", nil)
    }
  }
})

- Naren Krish