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请问，如果 iOS 设备上的屏幕打开或关闭，我如何使用 FireMonkey？

[英]Please, how I can get with FireMonkey if screen is on or off on iOS device?

 原文  2021-01-31 19:36:38  0   1    ios/ firemonkey

问题描述

I'm developing an App for iOS with Firemonkey (Delphi) that can run in background mode, but sometimes I need to know if device is in sleep mode, or with screen on or off (it also serves me).我正在为 iOS 开发一个可以在后台模式下运行的 Firemonkey (Delphi) 应用程序，但有时我需要知道设备是否处于睡眠模式，或者屏幕是否打开或关闭（它也为我服务）。

Please, any suggestions?.请问，有什么建议吗？ Thanks!谢谢！

1 个解决方案

解决方案1
 0  2021-01-31 22:29:33

From these constants you can get the App state (Active, Inactive and background)从这些常量中，您可以获得 App state（活动、非活动和后台）

https://developer.apple.com/documentation/uikit/uiapplication/state . https://developer.apple.com/documentation/uikit/uiapplication/state 。

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