通过 IOS 中的 Linking.openURL React Native Open 设置 [英] React Native Open settings through Linking.openURL in IOS

问题描述

我想从我的应用程序中打开 ios 设置应用程序.设置目标是 [设置 => 通知 => myapp].打开 &关闭推送通知.

I want to open ios setting app from my app. the settings destination is [ settings => notification => myapp ]. to turn on & turn off push notification.

关于如何链接到设置有一些文档，但我不知道如何打开深层链接.(通知 => myapp).

There are some documents about how to link to settings, but I don't know how to open deep link. (notification => myapp).

我该怎么做?

推荐答案

您可以像这样深入链接引用设置的索引:

You can deep-link referencing the settings's index like so:

Linking.openURL('app-settings:')

以上方法仅适用于IOS

Above method only for IOS

这篇关于通过 IOS 中的 Linking.openURL React Native Open 设置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！