在回调中以编程方式推送视图，SwiftUI [英] Push View programmatically in callback, SwiftUI

问题描述

在我看来，Apple 是在鼓励我们在 SwiftUI 中放弃使用 UIViewController，但是不使用视图控制器，我觉得有点无能为力.我想要的是能够实现某种ViewModel，它将向View 发出事件.

It seems to me that Apple is encouraging us to give up using UIViewController in SwiftUI, but without using view controllers, I feel a little bit powerless. What I would like is to be able to implement some sort of ViewModel which will emit events to View.

视图模型:

public protocol LoginViewModel: ViewModel {
  var onError: PassthroughSubject<Error, Never> { get }
  var onSuccessLogin: PassthroughSubject<Void, Never> { get }
}

查看:

public struct LoginView: View {
  fileprivate let viewModel: LoginViewModel
  
  public init(viewModel: LoginViewModel) {
    self.viewModel = viewModel
  }
  
  public var body: some View {
    NavigationView {
      MasterView()
        .onReceive(self.viewModel.onError, perform: self.handleError)
        .onReceive(self.viewModel.onSuccessLogin, perform: self.handleSuccessfullLogin)
    }
  }

  func handleSuccessfullLogin() {
    //push next screen
  }
  
  func handleError(_ error: Error) {
    //show alert
  }
}

使用SwiftUI，如果登录成功，我不知道如何推送另一个控制器

Using SwiftUI, I don't know how to push another controller if login is successful

此外，如果您有任何关于如何以更好的方式实现我想要的东西的建议，我将不胜感激.谢谢.

Also, I would appreciate any advice about how to implement what I want in a better way. Thanks.

推荐答案

我找到了答案.如果你想在回调中显示另一个视图，你应该

I've found the answer. If you want to show another view on callback you should

1. 创建状态 @State var pushActive = false

当 ViewModel 通知登录成功时，将 pushActive 设置为 true

When ViewModel notifies that login is successful set pushActive to true

func handleSuccessfullLogin() {
    self.pushActive = true
    print("handleSuccessfullLogin")
}

创建隐藏的NavigationLink并绑定到那个状态

NavigationLink(destination: 
   ProfileView(viewModel: ProfileViewModelImpl()),
   isActive: self.$pushActive) {
     EmptyView()
}.hidden()

这篇关于在回调中以编程方式推送视图，SwiftUI的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！