React Native Open settings through Linking.openURL in IOS

问题

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

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

How can I do this?

回答1:

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

Linking.openURL('app-settings:')

Above method only for IOS

回答2:

Since React Native 0.60 to open App settings use:

Linking.openSettings()

回答3:

Use Linking.openURL. For example, below is how to check and open Health app on iOS

import { Linking } from 'react-native'

async goToSettings() {
  const healthAppUrl = 'x-apple-health://'
  const canOpenHealthApp = await Linking.canOpenURL(healthAppUrl)
  if (canOpenHealthApp) {
    Linking.openURL(healthAppUrl)
  } else {
    Linking.openURL('app-settings:')
  }
}

回答4:

To access specific settings screens, try this: Linking.openURL("App-Prefs:root=WIFI"); Linking to app-settings only opens the settings for the Reference: iOS Launching Settings -> Restrictions URL Scheme (note that prefs changed to App-Prefs in iOS 6)

回答5:

Try this one for Open Specific System URL - Linking.openURL('App-Prefs:{3}')

回答6:

try this

Linking.openURL('app-settings://notification/myapp')

来源：https://stackoverflow.com/questions/44582694/react-native-open-settings-through-linking-openurl-in-ios