ios - [React Native][IOS] 如何通过通知打开应用程序时捕获事件

Question

React Native 有一个 PushNotificationIOS 模块来获取通知数据并监听 notificaiton、register 等事件。我想在通过通知打开应用程序时捕获一个事件:PushNotificationIOS.addEventListener('notification_open', myhandler);。有办法吗？

Answer 1

你有两个选择。

1。使用 PushNotificationIOS

您可以在重要的地方调用 PushNotificationIOS.getInitialNotification();。例如，您可以在应用的 index.js 文件中调用此方法并决定要呈现的组件。来自 the docs :

This method returns a promise. If the app was launched by a push notification, this promise resolves to an object of type PushNotificationIOS. Otherwise, it resolves to null.

2。使用原生 iOS App Delegate

来自 the UIApplicationDelegate documentation :

For example, if your app was launched because of an incoming remote notification, you might want to reconfigure your user interface to display data related to that notification. For a list of possible reasons why your app might be launched, see Launch Options Keys.

如果您出于某种原因需要提前知道，您可以检查 native application(_:willFinishLaunchingWithOptions:) 和 application(_:didFinishLaunchingWithOptions:) 方法。请注意，这些方法是在 React Native 初始化之前调用的。这是大多数 React Native 应用不需要的更高级的用例。