I'm trying to access my react-native app's specific permissions in Android. I can access and pull up my app's info page in Android, but I can't go directly into the permissions screen from my app.
I am using react-native-android-open-settings to to achieve this. By using this I am able to access the app's info page, but can't access the specific permissions page for the app without the user having to click on permissions in App Info. The library can be found here: https://www.npmjs.com/package/react-native-android-open-settings
import AndroidOpenSettings from 'react-native-android-open-settings'
async goToSettings() {
if(Platform.OS === 'android') {
AndroidOpenSettings.appDetailsSettings()
} else {
Linking.canOpenURL('app-settings:').then(supported => {
if (!supported) {
console.log('Can\'t handle settings url');
} else {
return Linking.openURL('app-settings:');
}
}).catch(err => console.error('An error occurred', err));
}
}
The expected result is for it to open the permissions page and not the app info page.
Related
I have a credit card addition system with 3DS security enabled.
To achieve that, I have to open a browser to the bank 3DS system.
As the user has the possibility to close the opened browser tab, I have to be able to detect this event and reset the loading state.
I did it that way:
client.service('credit-cards')
.patch(creditCardRef.current._id, {
registrationId: creditCardRef.current.registrationId,
registrationData: registrationDataRef.current,
registrationIpAddress: ipAddressRef.current,
// #ts-expect-error Unable to set the type of the parsed query.
registrationBrowserInfo: {
...parsedUrl.query,
acceptHeader: headersRef.current.accept,
},
secureModeRedirectUrl: Linking.createURL('paybox-3ds', {
queryParams: { creditCardId: creditCardRef.current._id },
}),
})
.then(ResultHelpers.toOne)
.then((result) => WebBrowser.openBrowserAsync(result.secureModeUrl, {
// #see https://github.com/expo/expo/issues/8072#issuecomment-621173298
showInRecents: true,
}))
.then((browserResult: WebBrowser.WebBrowserResult): void => {
// On IOS check if the user manually closed the browser using "Done" browser button
if (browserResult.type === 'cancel') {
setLoading(false);
}
})
.catch((error) => {
handleCreditCardRegistrationError(error);
});
This is working, but only for iOS. Indeed, the official documentation says:
The promise behaves differently based on the platform.
On Android promise resolves with {type: 'opened'} if we were able to
open browser.
On iOS: If the user closed the web browser, the Promise resolves with
{ type: 'cancel' }.
What is my alternative for Android?
I'd like for users to be able to share a link (e.g. app.com/SKFLA - this is primarily because deep links on their own aren't clickable) via Facebook etc. When clicked, this redirects to a deep link app://SKFLA. If the app is installed, this opens the app - this is all working fine so far. But if the app isn't installed, I'd like to open the app store on the relevant page. Is this achievable? Thanks!
You need UNIVERSAL LINKS
Please check
IOS https://developer.apple.com/library/archive/documentation/General/Conceptual/AppSearch/UniversalLinks.html
Android
https://developer.android.com/training/app-links/
It might also require some extra server-side setup.
Not sure about native behavior.
We used third-party service like https://branch.io/deepviews/.
There is a bunch of similar services.
If someone is still stuck in this issue and needs easiest solution, you will love node-deeplink
1.) If app is installed: Calling an app through deep linking will always call componentDidMount of root component. So you can attach a listener there. Like:
Linking.getInitialURL()
.then(url => {
if (url) {
this.handleOpenURL({ url });
}
})
.catch(console.error);
Linking.addEventListener('url', this.handleOpenURL);
handleOpenURL(event) {
if (event) {
console.log('event = ', event);
const url = event.url;
const route = url.replace(/.*?:\/\//g, '');
console.log('route = ', route);
if(route.match(/\/([^\/]+)\/?$/)) {
const id = route.match(/\/([^\/]+)\/?$/)[1];
const routeName = route.split('/')[0];
if (routeName === 'privatealbum') {
Actions.privateAlbum({ albumId: id });
}
}
}
}
2.) If app is not installed: Just set up a route in your server and node-deeplink package will handle the bridging between web browser to app store when a app is not installed in your mobile.
By this, both the cases will be handled without any struggle
I want to use Internet in app and need to ask for permission for both IOS/Android. I have checked out react-native-permissions but could not set up the module because of the react native link command. What is the easiest way to grant permissions in React-native? Any examples?
You can use internet by default if you use Expo (no need to ask internet permission). But if you want to ask the others permission check this link https://docs.expo.io/versions/latest/sdk/permissions.html
Example code :
async function getLocationAsync() {
const { Location, Permissions } = Expo;
const { status } = await Permissions.askAsync(Permissions.LOCATION);
if (status === 'granted') {
return Location.getCurrentPositionAsync({enableHighAccuracy: true});
} else {
throw new Error('Location permission not granted');
}
}
I have a button and I want to open a facebook page in the facebook app. I can use this solution to open the link in a browser but I'm looking for a better solution that opens faecbook app and my desire page. Is this generally possible? How?
This may not be possible on Android but to do so you follow essentially the same instructions for linking, you just need to swap out http with fb (or the appropriate app id). This SO answer has a bit more information on what may or may not be possible.
Assuming it is possible, to open the facebook app to a profile it would look something like this
const pageId = 'abc123'
Linking.openURL(`fb://profile/${pageId}`)
.catch(err => console.error('An error occurred', err));
Notice that rather than using http I'm using fb
Same as solution of #Spencer answered, but using page instead profile to open fanpage.
<Button
title="Go to Facebook page"
onPress={() => {
const FANPAGE_ID = 'xxxxxxxxxxxxxxxxx'
const FANPAGE_URL_FOR_APP = `fb://page/${FANPAGE_ID}`
const FANPAGE_URL_FOR_BROWSER = `https://fb.com/${FANPAGE_ID}`
Linking.canOpenURL(FANPAGE_URL_FOR_APP)
.then((supported) => {
if (!supported) {
Linking.openURL(FANPAGE_URL_FOR_BROWSER)
} else {
Linking.openURL(FANPAGE_URL_FOR_APP)
})
.catch(err => console.error('An error occurred', err))
}}
/>
Note: You MUST use fanpage ID, not fanpage slug name. If you don't know how to get id, just open your fanpage in browser, view source and find page_id param.
A mix of answers from #Spencer and #Thành worked for me on iOS.
So I settled for just attempting to open the Facebook app link, and then if that fails I fall back to the web browser link, like so:
import { Linking } from "react-native";
const openFacebookLink = facebookId => {
const FANPAGE_URL_FOR_APP = `fb://profile/${facebookId}`;
const FANPAGE_URL_FOR_BROWSER = `https://fb.com/${facebookId}`;
Linking.canOpenURL(FANPAGE_URL_FOR_APP)
.then(appSupported => {
if (appSupported) {
console.log(`Can handle native url: ${FANPAGE_URL_FOR_APP}`);
return Linking.openURL(FANPAGE_URL_FOR_APP);
} else {
console.log(
`Can't handle native url ${FANPAGE_URL_FOR_APP} defaulting to web URL ${FANPAGE_URL_FOR_BROWSER}`
);
return Linking.canOpenURL(FANPAGE_URL_FOR_BROWSER).then(
webSupported => {
if (webSupported) {
console.log(`Can handle web url: ${FANPAGE_URL_FOR_BROWSER}`);
return Linking.openURL(FANPAGE_URL_FOR_BROWSER);
}
return null;
}
);
}
})
.catch(err => console.error("An error occurred", err));
};
Note: the appSupported variable here will always return false until you've edited/added the LSApplicationQueriesSchemes value in your info.plist file. You'll find this file in the ios/yourappname sub-folder of your project. Here are the lines that I added to mine:
<key>LSApplicationQueriesSchemes</key>
<array>
<string>fb</string>
<string>fbapi</string>
<string>fb-messenger-api</string>
<string>fbauth2</string>
<string>fbshareextension</string>
</array>
NB: if you're using Create React Native App and/or Expo then you won't be able to edit this file. I abandoned Expo for this reason.
That works on iOS for me, but Android opens it in the browser every time. I've read that Android handles this stuff completely differently to iOS, so I'm not sure if there's any easy solution there.
How to open other apps (Gmail, Camera) from ReactNative. How can I pass data from current scene to other app?
I found this npm library react-native-app-link which can open other apps. This is based on deep linking, if you have any deep links then this library can help. This doesn't open apps just by giving the android package name or ios app id.
https://github.com/FiberJW/react-native-app-link
you can mange opening other apps using Linking
Code sample for opening the dialer
const urlToOpen = 'tel:1234567890';
Linking.openURL(urlToOpen);
You can refer to the official doc here, it just predefines some applications, which can be opened.
However, if the question is about to open just about any application, I hope there is no solution till now.
react-native-app-link has some redundant config (e.g. appStoreLocale parameter), so I wrote my own realization using their code:
import { Alert, Platform, ToastAndroid } from 'react-native';
const isIos = Platform.OS === 'ios';
const showNotification = (text) => isIos
? Alert.alert(text)
: ToastAndroid.show(text, ToastAndroid.SHORT);
const openApp = ({ url, appStoreId, playMarketId, name }) => {
Linking.openURL(url).catch(err => {
if (err.code === 'EUNSPECIFIED') {
Linking.openURL(
isIos
? `https://apps.apple.com/app/id${appStoreId}`
: `https://play.google.com/store/apps/details?id=${playMarketId}`,
);
} else {
showNotification(`Can't open ${name} app`);
}
});
};
It tries to open the app by the specified link, and if the user doesn't have such one, it opens its page in AppStore or Play Market.