如何执行push UIViewController或UIAlertController的礼物？

Question

After some API calls, I have a check to either navigate to another screen or show an alert on the same screen. Currently, I am doing this by creating an observable that returns a UIViewController type but pushing a UIAlertController causes problem.

Any suggestions/ideas on how this should be done?

ViewModel.swift

let nextAction = Observable.combineLatest(appVersionOutput, serviceAvailabilityOutput, getLanguagePackOutput,
                                              resultSelector:
        { appVersion, _, _ -> UIViewController in
            if appVersion.currentAppVersion == "1.0.0" {
                let appServiceAvailability = Availability.shared.getAppStatus()
                if appServiceAvailability {
                    return LoginLandingViewController()
                } else {
                    return ServiceMaintenanceViewController()
                }
            } else {
                return UIAlertController()
            }
        })

ViewController.swift

viewModel.output.nextAction
        .subscribe(onNext: { [weak self] screen in
            self?.navigationController?.pushViewController(screen, animated: true) 
        }) // PROBLEM FACED: Pushing a UIAlertController
        .disposed(by: disposeBag)

Answer 1

You can check the class using classForCoder

viewModel.output.nextAction
    .subscribe(onNext: { [weak self] screen in
        if String(describing: screen.classForCoder) == "UIAlertController" {
            //present
            self?.present(screen, animated: true, completion: nil)
        } else {
            //navigate
            self?.navigationController?.pushViewController(screen, animated: true)
        } 
    }) // PROBLEM FACED: Pushing a UIAlertController
    .disposed(by: disposeBag)

Answer 2

There're a few options.

The first option is the adoption of your current implementation. You could pass an information in addition to ViewController in onNext event of nextAction observable, that will tell how to show the VC.

For instance, you can create an associated enum

// you can call it NextAction, Action etc
enum PresentationType {
    case
    push(UIViewController),
    present(UIViewController)
}

and reuse it like this:

let nextAction = Observable.combineLatest(appVersionOutput, serviceAvailabilityOutput, getLanguagePackOutput,
                                          resultSelector:
    { appVersion, _, _ -> PresentationType in
        if appVersion.currentAppVersion == "1.0.0" {
            let appServiceAvailability = Availability.shared.getAppStatus()
            if appServiceAvailability {
                return .push(LoginLandingViewController())
            } else {
                return .push(ServiceMaintenanceViewController()) // use .present if should present modally
            }
        } else {
            return .present(UIAlertController())
        }
})

// somewhere in viewController

viewModel.output.nextAction
        .subscribe(onNext: { [weak self] action in
        switch action {
            case .push(let vc):
                self?.navigationController?.pushViewController(vc, animated: true)
            case .present(let vc):
                self?.present(vc, animated: true, completion: nil)

        })
        .disposed(by: disposeBag)

The second (and more flexible and testable in some sense) option would be to create a separate Router class which is responsible for creating and showing next screens (with functions showLogin , showAlert etc). The router can be directly injected in ViewModel and you can call the Router to show next screens in, for example, do(onNext) events in your observables.

Answer 3

You can look at the is keyword which allows you to check the object type. More on is keyword . An alternative will be to use type(of: object) and compare to UIAlertViewController.self

Since you have to present a UIAlertViewController and not push it, use if and else with the above to present if type is UIAlertViewController and push if otherwise. NB: Unnecessary to check for UIViewController since they all are.