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User object contains nil after POST request to server


How to detect tableView cell touched or clicked in swiftFirebase swift ios login system error Assertion failed/Exec_BAD_INSTRUNCTION (code=EXC_i386_INVOP, subcode=0x0)Expand and Collapse tableview cellsTableView not displaying text with JSON data from API callHow to show images from API in CollectionViewAdding a custom UIViewcontroller to subview programmatically but getting an error message “Cannot convert value of type…”How to optimize UITableViewCell, because my UITableView lagsSwift Error - Use of undeclared type 'cell' - Collection ViewTableView Controller is hiding the dropshadow of swipemenu in swift 4BMI app print result is 0 even with variable being hard coded






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








-2















After sending a POST request using alamofire to register a user in my app, I am attempting to create a User object from the returned user_id from my server. However, when I attempt to print some attributes of this user object directly after I initialize it, it works fine. However, when I try and print some attributes of it after the alamofire post request has finished executing, I am getting nil.



import UIKit
import Alamofire

class ViewController: UIViewController
var user: User!
@IBOutlet weak var nameField: UITextField!
@IBOutlet weak var emailField: UITextField!
@IBOutlet weak var phone_number_field: UITextField!

override func viewDidLoad()
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.


//MARK: Actions
@IBAction func generateQR(_ sender: UIButton)
let name = nameField.text
let email = emailField.text
let phone_number = phone_number_field.text
let params: Parameters = ["name": name!, "email": email!, "phone_number": phone_number!]
AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
switch response.result
case .success:
let user_id = response.result.value
self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)
print(self.user?.name) // <--This print statement works correctly -->
case .failure:
print("Error")


print(self.user?.name) // <--This prints nil and in the debugger, self.user shows as nil-->
performSegue(withIdentifier: "qrSegue", sender: self)


//END Actions

override func prepare(for segue: UIStoryboardSegue, sender: Any?)
if segue.destination is QRCodeController
//Pass user object
let qr_controller = segue.destination as? QRCodeController
qr_controller?.user = self.user













share|improve this question






















  • performSegue call first then after completion block call. you have to add segue code into success case.

    – Vishal Patel
    Mar 26 at 5:22











  • Add print("Alamofire Completion") just before the line switch response.result. You'll see that the order of print(self.user?.name) output and that one is not the one you think of. You are missing the asynchrone concept.

    – Larme
    Mar 26 at 11:19

















-2















After sending a POST request using alamofire to register a user in my app, I am attempting to create a User object from the returned user_id from my server. However, when I attempt to print some attributes of this user object directly after I initialize it, it works fine. However, when I try and print some attributes of it after the alamofire post request has finished executing, I am getting nil.



import UIKit
import Alamofire

class ViewController: UIViewController
var user: User!
@IBOutlet weak var nameField: UITextField!
@IBOutlet weak var emailField: UITextField!
@IBOutlet weak var phone_number_field: UITextField!

override func viewDidLoad()
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.


//MARK: Actions
@IBAction func generateQR(_ sender: UIButton)
let name = nameField.text
let email = emailField.text
let phone_number = phone_number_field.text
let params: Parameters = ["name": name!, "email": email!, "phone_number": phone_number!]
AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
switch response.result
case .success:
let user_id = response.result.value
self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)
print(self.user?.name) // <--This print statement works correctly -->
case .failure:
print("Error")


print(self.user?.name) // <--This prints nil and in the debugger, self.user shows as nil-->
performSegue(withIdentifier: "qrSegue", sender: self)


//END Actions

override func prepare(for segue: UIStoryboardSegue, sender: Any?)
if segue.destination is QRCodeController
//Pass user object
let qr_controller = segue.destination as? QRCodeController
qr_controller?.user = self.user













share|improve this question






















  • performSegue call first then after completion block call. you have to add segue code into success case.

    – Vishal Patel
    Mar 26 at 5:22











  • Add print("Alamofire Completion") just before the line switch response.result. You'll see that the order of print(self.user?.name) output and that one is not the one you think of. You are missing the asynchrone concept.

    – Larme
    Mar 26 at 11:19













-2












-2








-2








After sending a POST request using alamofire to register a user in my app, I am attempting to create a User object from the returned user_id from my server. However, when I attempt to print some attributes of this user object directly after I initialize it, it works fine. However, when I try and print some attributes of it after the alamofire post request has finished executing, I am getting nil.



import UIKit
import Alamofire

class ViewController: UIViewController
var user: User!
@IBOutlet weak var nameField: UITextField!
@IBOutlet weak var emailField: UITextField!
@IBOutlet weak var phone_number_field: UITextField!

override func viewDidLoad()
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.


//MARK: Actions
@IBAction func generateQR(_ sender: UIButton)
let name = nameField.text
let email = emailField.text
let phone_number = phone_number_field.text
let params: Parameters = ["name": name!, "email": email!, "phone_number": phone_number!]
AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
switch response.result
case .success:
let user_id = response.result.value
self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)
print(self.user?.name) // <--This print statement works correctly -->
case .failure:
print("Error")


print(self.user?.name) // <--This prints nil and in the debugger, self.user shows as nil-->
performSegue(withIdentifier: "qrSegue", sender: self)


//END Actions

override func prepare(for segue: UIStoryboardSegue, sender: Any?)
if segue.destination is QRCodeController
//Pass user object
let qr_controller = segue.destination as? QRCodeController
qr_controller?.user = self.user













share|improve this question














After sending a POST request using alamofire to register a user in my app, I am attempting to create a User object from the returned user_id from my server. However, when I attempt to print some attributes of this user object directly after I initialize it, it works fine. However, when I try and print some attributes of it after the alamofire post request has finished executing, I am getting nil.



import UIKit
import Alamofire

class ViewController: UIViewController
var user: User!
@IBOutlet weak var nameField: UITextField!
@IBOutlet weak var emailField: UITextField!
@IBOutlet weak var phone_number_field: UITextField!

override func viewDidLoad()
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.


//MARK: Actions
@IBAction func generateQR(_ sender: UIButton)
let name = nameField.text
let email = emailField.text
let phone_number = phone_number_field.text
let params: Parameters = ["name": name!, "email": email!, "phone_number": phone_number!]
AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
switch response.result
case .success:
let user_id = response.result.value
self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)
print(self.user?.name) // <--This print statement works correctly -->
case .failure:
print("Error")


print(self.user?.name) // <--This prints nil and in the debugger, self.user shows as nil-->
performSegue(withIdentifier: "qrSegue", sender: self)


//END Actions

override func prepare(for segue: UIStoryboardSegue, sender: Any?)
if segue.destination is QRCodeController
//Pass user object
let qr_controller = segue.destination as? QRCodeController
qr_controller?.user = self.user










swift xcode alamofire






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 26 at 5:12









nilay neeranjunnilay neeranjun

7011 bronze badges




7011 bronze badges












  • performSegue call first then after completion block call. you have to add segue code into success case.

    – Vishal Patel
    Mar 26 at 5:22











  • Add print("Alamofire Completion") just before the line switch response.result. You'll see that the order of print(self.user?.name) output and that one is not the one you think of. You are missing the asynchrone concept.

    – Larme
    Mar 26 at 11:19

















  • performSegue call first then after completion block call. you have to add segue code into success case.

    – Vishal Patel
    Mar 26 at 5:22











  • Add print("Alamofire Completion") just before the line switch response.result. You'll see that the order of print(self.user?.name) output and that one is not the one you think of. You are missing the asynchrone concept.

    – Larme
    Mar 26 at 11:19
















performSegue call first then after completion block call. you have to add segue code into success case.

– Vishal Patel
Mar 26 at 5:22





performSegue call first then after completion block call. you have to add segue code into success case.

– Vishal Patel
Mar 26 at 5:22













Add print("Alamofire Completion") just before the line switch response.result. You'll see that the order of print(self.user?.name) output and that one is not the one you think of. You are missing the asynchrone concept.

– Larme
Mar 26 at 11:19





Add print("Alamofire Completion") just before the line switch response.result. You'll see that the order of print(self.user?.name) output and that one is not the one you think of. You are missing the asynchrone concept.

– Larme
Mar 26 at 11:19












1 Answer
1






active

oldest

votes


















1














Your service call is async. Read about sync vs async methods.



Medium



Ray Wenderlich



AF sent request to the server and codes below continues working. When your response will arrive then success or failure blocks start working.



So the user object is not assigned until the success block is running.




Here is solution:




 AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
switch response.result
case .success:
let user_id = response.result.value
self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)

print(self.user?.name) // <--This print statement works correctly -->
performSegue(withIdentifier: "qrSegue", sender: self)

case .failure:
print("Error")




And you can show loading in your page while pending response from request and hide it on success.






share|improve this answer
























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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Your service call is async. Read about sync vs async methods.



    Medium



    Ray Wenderlich



    AF sent request to the server and codes below continues working. When your response will arrive then success or failure blocks start working.



    So the user object is not assigned until the success block is running.




    Here is solution:




     AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
    switch response.result
    case .success:
    let user_id = response.result.value
    self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)

    print(self.user?.name) // <--This print statement works correctly -->
    performSegue(withIdentifier: "qrSegue", sender: self)

    case .failure:
    print("Error")




    And you can show loading in your page while pending response from request and hide it on success.






    share|improve this answer





























      1














      Your service call is async. Read about sync vs async methods.



      Medium



      Ray Wenderlich



      AF sent request to the server and codes below continues working. When your response will arrive then success or failure blocks start working.



      So the user object is not assigned until the success block is running.




      Here is solution:




       AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
      switch response.result
      case .success:
      let user_id = response.result.value
      self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)

      print(self.user?.name) // <--This print statement works correctly -->
      performSegue(withIdentifier: "qrSegue", sender: self)

      case .failure:
      print("Error")




      And you can show loading in your page while pending response from request and hide it on success.






      share|improve this answer



























        1












        1








        1







        Your service call is async. Read about sync vs async methods.



        Medium



        Ray Wenderlich



        AF sent request to the server and codes below continues working. When your response will arrive then success or failure blocks start working.



        So the user object is not assigned until the success block is running.




        Here is solution:




         AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
        switch response.result
        case .success:
        let user_id = response.result.value
        self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)

        print(self.user?.name) // <--This print statement works correctly -->
        performSegue(withIdentifier: "qrSegue", sender: self)

        case .failure:
        print("Error")




        And you can show loading in your page while pending response from request and hide it on success.






        share|improve this answer















        Your service call is async. Read about sync vs async methods.



        Medium



        Ray Wenderlich



        AF sent request to the server and codes below continues working. When your response will arrive then success or failure blocks start working.



        So the user object is not assigned until the success block is running.




        Here is solution:




         AF.request("http://127.0.0.1:5000/register", method: .post, parameters: params, encoding: JSONEncoding.default).responseString response in
        switch response.result
        case .success:
        let user_id = response.result.value
        self.user = User(name: name!, email: email!, phone_number: phone_number!, user_id: user_id!)

        print(self.user?.name) // <--This print statement works correctly -->
        performSegue(withIdentifier: "qrSegue", sender: self)

        case .failure:
        print("Error")




        And you can show loading in your page while pending response from request and hide it on success.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 26 at 5:23

























        answered Mar 26 at 5:17









        ibrahimyilmazibrahimyilmaz

        1,56315 silver badges25 bronze badges




        1,56315 silver badges25 bronze badges


















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