How to go back to previous screen after a timeout request?

Question

I'm trying to set up the network session creator to go back to the previous screen if the timeout request runs out. As of now I'm not completely sure where or how to execute it. Here is the code:

  lazy var configuration: URLSessionConfiguration = URLSessionConfiguration.default

  lazy var session: URLSession = URLSession(configuration: self.configuration)

 typealias JSONData = ((Data) -> Void)

  func getJSONData(type: String, urlExtension: String, completion: @escaping JSONData) {

    configuration.timeoutIntervalForRequest = 5
    configuration.timeoutIntervalForResource = 5

    let request = URLRequest(url: URL(string:"\(baseURL)\(type)/\(urlExtension)?api_key=\(apiKey)")! )

    let dataTask = session.dataTask(with: request, completionHandler: { (data, response, error) in

      if error == nil {
        if let httpResponse = response as? HTTPURLResponse {
          switch (httpResponse.statusCode) {
          case 200:
            if let data = data {
              completion(data)
            }
          default:
            print(httpResponse.statusCode)
          }
        }
      } else {
        print("Error: \(error?.localizedDescription)")
      }
    })
    dataTask.resume()
  }

Are you using UINavigationController? – Qadir Hussain Mar 14 '17 at 14:00 — Qadir Hussain, Mar 14 '17 at 14:00
I am using a UINavigationController – SwiftyJD Mar 14 '17 at 14:01 — SwiftyJD, Mar 14 '17 at 14:01

unniverzal · Accepted Answer · 2017-03-14T14:23:43.373

 lazy var configuration: URLSessionConfiguration = URLSessionConfiguration.default

  lazy var session: URLSession = URLSession(configuration: self.configuration)

 typealias JSONData = ((Data) -> Void)

  func getJSONData(type: String, urlExtension: String, onSucceed : @escaping JSONData ,  onFailure: @escaping (_ error:NSError)-> Void ) {

    configuration.timeoutIntervalForRequest = 5
    configuration.timeoutIntervalForResource = 5

    let request = URLRequest(url: URL(string:"\(baseURL)\(type)/\(urlExtension)?api_key=\(apiKey)")! )

    let dataTask = session.dataTask(with: request, completionHandler: { (data, response, error) in

      if error == nil {
        if let httpResponse = response as? HTTPURLResponse {
          switch (httpResponse.statusCode) {
          case 200:
            if let data = data {
              onSucceed(data)
            }
          default:
            print(httpResponse.statusCode)
          }
        }
      } else {
        onFailure(error! as NSError)
      }
    })
    dataTask.resume()
  }

In the View Controller where you call this method I would use it like this.

NetworkManager.getJSONData(type: "", urlExtension: "",onSucceed {
   //doSmth
}, on Failure{(error) in 
   //show Error
// if it is pushed
    _ = self.navigationController?.popViewController(animated: true) 

// or if its presented 
// self.navigationController?.dismiss(animated: true, completion: nil)

}

This request is made inside a singleton but is accessed by the viewcontrollers — SwiftyJD, Mar 14 '17 at 14:09
I edited my answer , I added 2 closures : one if the request succeeds and the other one if it fails. Depending on the case you can manage them both. — unniverzal, Mar 14 '17 at 14:25

score 0 · Answer 2 · answered Mar 14 '17 at 14:06

0

You can try this. this is working for me

_ = self.navigationController?.popViewController(animated: true)

this will pop the top most UIViewController from the UINavigationController stack, as a result your previous screen will be visible.

answered Mar 14 '17 at 14:06

Qadir Hussain

8,721
13
89
124

score 0 · Answer 3 · answered Mar 14 '17 at 14:07

0

If you used a show segue to get to your current VC you can use:

_ = self.navigationController?.popViewController(animated: true)

If you used a modal segue you would use:

_ = self.navigationController?.dismiss(animated: true)

answered Mar 14 '17 at 14:07

Casper Zandbergen

3,419
2
25
49

How to go back to previous screen after a timeout request?

3 Answers3