How to lock screen orientation for a specific view?

Question

Created Feb ’20

Replies 2

Boosts 0

Views 9.9k

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I have one view, that need to be locked in portrait only on the iPhone, but not on the iPad, but while it seems to be kinda easy on UIKit, I cannot find any solution for the SwiftUI. For UIKit it works through overriding supportedInterfaceOrientations in the controller, but I don't see any easy way to do so in the SwiftUI

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Answer 1

Jim Dovey OP

Feb ’20

Oooh, I know this one! There's a discussion here: SwiftUI: Force orientation on a per screen basis.

I also implemented this as part of my open-source toolkit: https://github.com/AlanQuatermain/AQUI/blob/master/Sources/AQUI/OrientationLock.swift

The version on Github allows views to specify their individual constraints, and will collapse them all into a single allowable state. If one view says 'all but upside down' and another says 'portrait only' then the effective value will be 'portrait only'.

0

Answer 2

birdmichael OP

Aug ’20

struct NetflixApp: App {

  @UIApplicationDelegateAdaptor(AppDelegate.self) var appDelegate

}

class AppDelegate: NSObject {

  // 屏幕旋转锁定

  static var orientationLock = UIInterfaceOrientationMask.portrait

}

extension AppDelegate: UIApplicationDelegate {

  func application(_ application: UIApplication, supportedInterfaceOrientationsFor window: UIWindow?) -> UIInterfaceOrientationMask {

    return AppDelegate.orientationLock

  }

}

SomeView
.onDisappear(perform: {

      DispatchQueue.main.async {

        AppDelegate.orientationLock = UIInterfaceOrientationMask.portrait

        UIDevice.current.setValue(UIInterfaceOrientation.portrait.rawValue, forKey: "orientation")

        UINavigationController.attemptRotationToDeviceOrientation()

      }

    })

    .onAppear(perform: {

      AppDelegate.orientationLock = UIInterfaceOrientationMask.landscape

      UIDevice.current.setValue(UIInterfaceOrientation.landscapeLeft.rawValue, forKey: "orientation")

      UINavigationController.attemptRotationToDeviceOrientation()

    })

2