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?
You can deep-link referencing the settings's index like so:
Linking.openURL('app-settings:')
Above method only for IOS
Since React Native 0.60 to open App settings use:
import { Linking } from 'react-native';
Linking.openSettings();
Open the app’s custom settings, if it has any.
Works for Android and iOS
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:')
}
}
for iOS 14, this is how i open location service settings
Linking.openURL('App-Prefs:Privacy&path=LOCATION')
tested in react native 0.63.4
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)
Try this one for Open Specific System URL - Linking.openURL('App-Prefs:{3}')
try this
Linking.openURL('app-settings://notification/myapp')
Related
I am using Ionic React / React Native to develop a mobile app. I want to use Leaflet to display the static map and markers. If the user click the markers, I want to open the coordinate in Google Map / Waze / any native map application then use that native app to navigate the user into the marker.
Is there a way to open the marked place into native map apps?
Google Map provides cross-platform deep linking universal URLs. If the native Google Map app is installed. It will open with that app, otherwise, use the default web browser installed on user's device.
You can use React Native Linking API to dynamically link to a specific location.
import { Linking, Alert } from "react-native";
async function openWithGoogleMap() {
/** Google Map API offers different URL to deep link to various locatioon and map ressources.
* Check their docs - https://developers.google.com/maps/documentation/urls/get-started
*/
const url =
"https://www.google.com/maps/search/?api=1&query=47.5951518%2C-122.3316393";
const supported = await Linking.canOpenURL(url);
if (supported) {
// Opening the link with some app, if the URL scheme is "http" the web link should be opened
// by some browser in the mobile
await Linking.openURL(url);
} else {
Alert.alert(`Don't know how to open this URL: ${url}`);
}
}
I am creating an APP using managed expo react native.
And want to implement Toast alerts, react-native provides Toast only for Android not for IOS.
I googled it and found the couple of modules which works on Android and ios but they required some config change in Native code. But as I said I am working on Managed expo app. So, I don't have access for that.
Now let me know how I can implement Toast on this?
Thanks.
As toast is a native feature in android, for ios try snakbar.
import {
ToastAndroid,
Platform
} from "react-native";
import Snackbar from 'react-native-snackbar';
notify = (message) => {
if (Platform.OS != 'android') {
Snackbar.show({
text: message,
duration: Snackbar.LENGTH_SHORT,
});
} else {
ToastAndroid.show(message, ToastAndroid.SHORT);
}
}
** If you are on expo
https://snack.expo.io/#mainak/snackbar
You can use third-party library native-base available for both react-native-cli and expo
[Native-Base] https://docs.nativebase.io/docs/GetStarted.html
[Toast Component] https://docs.nativebase.io/Components.html#toast-def-headref
I have developed a mobile application in react-native and on click of a icon I want to open the Qliksense app which is installed in mobile
Use Linking and the desired app url scheme:
import Linking:
import {Linking} from 'react-native';
Then on your function/action you would like to open the other app:
const canOpenUrl = await Linking.canOpenURL(url);
if (canOcanOpenUrl) {
Linking.openURL(url);
}
See more in the docs
I am new to using code-push for react native app, Everything works okay however I want to hide the Ignore button in the dialog. I dont want the user to ignore any update I have provided, Is this possible?
From the command line, you can use the mandatory option to hide the ignore button:
# Release a mandatory update with a changelog
code-push release-react MyApp-iOS ios -m --description "Modified the header color"
If you don't want user ignore the update, then don't show the dialog at all.
Just change codePushOptions so the user always get update when open the app from background in next time:
import React from 'react';
import {
AppRegistry,
} from 'react-native';
import codePush from "react-native-code-push";
let codePushOptions = {
checkFrequency: codePush.CheckFrequency.ON_APP_RESUME,
installMode: codePush.InstallMode.ON_NEXT_RESUME,
};
import App from './app/app';
AppRegistry.registerComponent('yourApp', () => codePush(codePushOptions)(App));
To begin with it is not recommended to use UpdateDialog within your application because the App could be rejected by AppStore at all due to AppStore Review Guidelines states that
Apps must not force users to rate the app, review the app, download
other apps, or other similar actions in order to access functionality,
content, or use of the app.
So use it with care and also please see this reference for more details.
As for hiding Ignore button - you could do it providing an empty string instead of button name e.g. as follows:
CodePush.sync(
{ installMode: CodePush.InstallMode.IMMEDIATE, updateDialog: { optionalIgnoreButtonLabel: "" } }, null, null );
you can also do the same thing with the help of appcenter webview page .
you just only need to:
go into the appcenter and indside your AppOverview>>Distribute>>CodePush: .
click on setting icon and click ON on the update required btn.
I'm trying to link with android os settings page.
Going to Phone Dail with Linking is work.
Linking.openURL('tel: +959 XXXXXXXX');
Can I use Linking to link with android os settings page like android location settings page?
Linking.openURL can only open correctly formatted urls. There are no standardized urls for the android settings. You would have to write some native code to do this, and then build a React Native Module to trigger it.
Here is some example Android code:
public void openWifiSettings() {
Intent intent = new Intent(Intent.ACTION_WIFI_SETTINGS);
if (intent.resolveActivity(getPackageManager()) != null) {
startActivity(intent);
}
}