Deep linking launches the app, but handles the external url wrongly - react-native

I am developing an React Native project, using React-Navigation v5.
I now try to implement deep linking feature. I followed the official instruction set up deep linking successfully (I mean the app is launched by custom url scheme). Next I need to handle the external link, my issue comes at this point.
To handle external link in my react-native project, I also followed the instruction configure links.
I defined a linking.js in my project:
const config = {
screens: {
// I explained why I nest FooScreen like this if you continue reading
FeatureFlow: {
SubfeatureFlow: {
FooScreen: {
path: 'foo/:myId',
},
},
},
},
};
const linking = {
prefixes: ['myapp://mobile'],
config,
};
export default linking;
Then, in my NavigationContainer, I use the linking as below:
return (
<NavigationContainer
linking={linking}
...
>
<MainFlow />
</NavigationContainer>
As you can see above, three things worth a notice:
in linking.js, inside config I specified that the path e.g. foo/123 should open screen FooScreen.
The FooScreen is a nested screen.
NavigationContainer contains a component called MainFlow.
To illustrated how FooScreen is nested in navigation hierarchy, let's start with MainFlow, which looks like this:
const MainFlow = ({navigation}) => {
const Drawer = createDrawerNavigator();
return (
<Drawer.Navigator
...>
<Drawer.Screen name="FeatureFlow" component={MyFeatureFlow} />
...
</Drawer.Navigator>
);
};
As you can see, MainFlow is a DrawerNavigator which hosts a screen named FeatureFlow refers to component MyFeatureFlow.
And MyFeatureFlow looks like this:
const MyFeatureFlow = ({navigation}) => {
const FeatureStack = createStackNavigator();
return (
<FeatureStack.Navigator
...>
<FeatureStack.Screen
name="SubfeatureFlow"
component={MySubfeatureFlow}/>
</FeatureStack.Navigator>
)
As you can see above, FeatureFlow is a stack navigator which hosts a screen named SubfeatureFlow, refers to component MySubfeatureFlow.
And MySubfeatureFlow is like:
const MySubfeatureFlow = ({navigation}) => {
const SubfeatureStack = createStackNavigator();
return (
<SubfeatureStack.Navigator
...>
<SubfeatureStack.Screen
name="FooScreen"
component={MyFooScreen}
</SubfeatureStack.Navigator>
)
As you can see above, MySubfeatureFlow is another stack navigator which hosts a screen named FooScreen refers to component MyFooScreen.
Now you understand why in linking.js configuration, the FooScreen is nested like that.
Then, I run my app on iOS simulator. The app launched. I killed the app on simulator.
Then, I run command npx uri-scheme open myapp://mobile/foo/123 --ios. (I also tried open mobile browser, type in url myapp://mobile/foo/123, same error as below)
I see on simulator, the app is launched by the command, which means my deep linking is set up. However, handling external link to open FooScreen fails, I end up with the following error:
The error tells me the navigation payload tries to treat 123 as screen, and the app treat foo as name of the screen in my project. I completely don't understand why it is. Is there something wrong with my linking configuration?

Related

React-native-web Multi-Platform Setup using Expo 44 + Typescript

What is the simplest way to implement Multi-Platform Setup for a component in Expo. I have tried mamy diferent ways.. it was working on web but it is failing on Native and failing with Jest & #testing-library/react-native. Ideally I would like the least amount of custom config etc (do not want to eject). I expect the file structure to look like this:
Component
|- index.tsx
|- Component.native.tsx
|- Component.web.tsx
I am not sure how to do the index.tsx. I saw someone say something like this would work:
// index.tsx
// #ts-ignore
export { default } from "Component"
this didn't work so I did
// index.tsx
// #ts-ignore
export { default } from "./Component"
This worked for web, but the jest test said
Cannot find './Component'
However, Jest was able to find:
'./Component.mobile.tsx'
'./Component.web.tsx'
I tried:
// index.tsx
// #ts-ignore
import Component from "./Component";
export default Component
and the tests was the same
and the native emulator said:
Unable to resolve module ./Component
I tried using lazy loading but this does not work on web.
import { lazy, Suspense } from "react";
import { Platform } from "react-native";
import Loading from "../../components/Loading";
import { ComponentType } from "./types";
const Web = lazy(() => import("./Component.web"));
const Mobile = lazy(() => import("./Component.mobile"));
const Component: ComponentType = (props) => {
const isWeb = Platform.OS === "web";
return (
<Suspense fallback={<Loading message="Loading Component" />}>
{isWeb ? <Web {...props} /> : <Mobile {...props} />}
</Suspense>
);
};
export default Component
Questions
how to use diferent files for components depending on platform (exlude other files from build)
how to make it ok with ts in vscode
Using Expo 44. Thanks
I would use named exports. So begin by having the same component name in both files. Next I would have one file called Component.tsx and the other Component.native.tsx. That will be enough to allow the bundler to pull the native for native and the other for non-native (in other words web). That should be all you need.

Expo deep linking in React Native Web

Trying to make deep linking for React Native Web work. Browsed many articles and finally found a solution proposed for the exact problem faced in this link whch doesn't work for me.
Expo documentation says use
const prefix = Linking.makeUrl('/')
and set the prefix in the AppNavigator.
Wondering how can I point to my react native web app when use clicks a link. The problem is Linking.makeUrl('/') returns exp://localhost:19006/ on my laptop. As browser wouldn't recognize exp as scheme is unable to open my URL.
I tried setting the scheme to https in app.json but still Linking.makeUrl('/') returns the scheme as expo.
In my app when user clicks on static html page which I load in the app I want to route back to my app. As I am unable to get right scheme for React Native Web using Expo unable to proceed further.
Thank you.
Attaching the code provided in the link.
import React from 'react'
import { createSwitchNavigator } from '#react-navigation/core'
import { createBrowserApp } from "#react-navigation/web"
import { Linking } from 'expo'
const prefix = Linking.makeUrl('/')
const AppNavigator = createBrowserApp(createSwitchNavigator({
Landing: {
screen: LandingScreen,
path: '/'
},
Login: {
screen: LoginScreen,
path: 'login'
}
}, {
initialRouteName: 'Landing',
backBehavior: 'history'
}))
export default () => <AppNavigator uriPrefix={prefix} />
Try :
const prefix = Linking.createURL('/');

My React Native app crashes during Login and Logout ( when context is changed )

I have a react native app and whenever I try to login or logout the app crashes .. I am not getting any error logs in React Native Debugger.
so the App Crashes and then when I reopen the app the changes have already taken effect ( i.e :- If I was trying to login I would be logged in when I reopen the app)
I have 2 files for Navigation
CofficNavigator.tsx
// This file has all the routing
// here BottomNavigator is a tab Navigator and AuthNavigator is a stack navigator
// Before the app crashing all the correct state is set but it just crashes
// ----------- Final Stack Navigator ----------------
const FinalStackNavigator = createStackNavigator();
export const FinalNavigator = (props: any) => {
const authContext = useContext(AuthContext);
const [signedIn, setSignedIn] = useState<boolean | null>(
authContext.isLoggedIn
);
useEffect(() => {
if (authContext.isLoggedIn !== signedIn && !authContext.isLoading) {
setSignedIn(authContext.isLoggedIn);
}
}, [authContext]);
return (
<FinalStackNavigator.Navigator>
{signedIn ? (
<>
<FinalStackNavigator.Screen
name="BottomNavigator"
component={BottomNavigator}
options={{ headerShown: false }}
/>
</>
) : (
<>
<FinalStackNavigator.Screen
name="AuthNavigator"
component={AuthNavigator}
/>
</>
)}
</FinalStackNavigator.Navigator>
);
};
AppNavigator.tsx
export const AppNavigator = (props: any) => {
return (
<NavigationContainer>
<FinalNavigator />
</NavigationContainer>
);
};
-------- Login Logic flow -------------
on success of login API I call this line
await authContext.changeCofficToken(result.data.login.token);
AuthContext.ts
const changeCofficToken = async (token: string) => {
await setItem("cofficToken", token);
setIsLoggedIn(true);
setCofficToken(token);
};
helper.ts
export const setItem = async (key: string, value: any) => {
await AsyncStorage.setItem(key, value);
};
App.tsx
return (
<AuthContextProvider>
<FilterContextProvider>
<SearchTermContextProvider>
<ShowVerticalListContextProvider>
<CouponContextProvider>
<ApolloProvider client={client}>
<AppNavigator />
</ApolloProvider>
</CouponContextProvider>
</ShowVerticalListContextProvider>
</SearchTermContextProvider>
</FilterContextProvider>
</AuthContextProvider>
);
------ End of Login Logic code flow -------
Error logs from my IOS Simulator
com.apple.CoreSimulator.SimDevice.957FA264-501C-44C3-80F3-9E8D1F600A1E[15500] (UIKitApplication:host.exp.Exponent[e036][rb-legacy][15601]): Service exited due to SIGABRT
assertion failed: 19F101 17F61: libxpc.dylib + 83746 [ED46009E-B942-37CC-95D4-82CF3FE90BD8]: 0x7d
Please Note :- I am using Expo if it matters
Video Link :- https://fisico-dhaval.s3.ap-south-1.amazonaws.com/Screen+Recording+2020-08-20+at+5.38.06+PM.mov
If you need any other code snippets then please do lemme know
Any help will be appreciated
Can you share the code where you setSignin. In case if you are saving and clearing token, make sure those libraries are supported and linked properly.
Service exited due to SIGABRT is a common issue in Native iOS development, it can be cause due to various events. While working on react-native, I have never experienced this error. But I'm adding some of my observation, may be that can help you,
This can be cause due to some buggy component, which are not compatible with Expo version. So, you need to add many console.log or may be comment some things and check is there any functionality or a component causing this error. A buggy component can be a culprit.
If you have pod installed, Make sure you check all your libraries.You likely have some issue with an installed external library. and make sure you run pod install and delete node_modules folder and then run npm install again (close the packager too)
You can also use sentry or any other crash reporting library to detect crashes, It will give you the exact line number.
Sometimes the crash reporting libraries won't work, the reason can be that the crash occurs before the js bundle is loaded and because of that, the library is not yet initialised and it can't detect the crash. If so you can install a native crash reporting library, You can use fabric for that, or even bugsnag.
These are few scenarios, I've observed via which you can debug or detect what's the real cause of this error.

How to setup sendSignInLinkToEmail() from Firebase in react-native?

Working on a react-native project using #react-native-firebase/app v6 we recently integrated signing in with a "magic link" using auth.sendSignInLinkToEmail
We couldn't find a good example on how to setup everything in react-native and had different problems like
- auth/invalid-dynamic-link-domain - The provided dynamic link domain is not configured or authorized for the current project.
- auth/unauthorized-continue-uri - Domain not whitelisted by project
Searching for information and implementing the "magic link login" I've prepared a guide on how to have this setup in react-native
Firebase project configuration
Open the Firebase console
Prepare firebase instance (Email Link sign-in)
open the Auth section.
On the Sign in method tab, enable the Email/Password provider. Note that email/password sign-in must be enabled to use email link sign-in.
In the same section, enable Email link (passwordless sign-in) sign-in method.
On the Authorized domains tab (just bellow)
Add any domains that will be used
For example the domain for the url from ActionCodeSettings needs to be included here
Configuring Firebase Dynamic Links
For IOS - you need to have an ios app configured - Add an app or specify the following throughout the firebase console
Bundle ID
App Store ID
Apple Developer Team ID
For Android - you just need to have an Android app configured with a package name
Enable Firebase Dynamic Links - open the Dynamic Links section
“Firebase Auth uses Firebase Dynamic Links when sending a link that is meant to be opened in a mobile application.
In order to use this feature, Dynamic Links need to be configured in the Firebase Console.”
(ios only) You can verify that your Firebase project is properly configured to use Dynamic Links in your iOS app by opening
the following URL: https://your_dynamic_links_domain/apple-app-site-association
It should show something like:
{
"applinks": {
"apps": [],
"details": [
{
"appID": "AP_ID123.com.example.app",
"paths": [
"NOT /_/", "/"
]
}
]
}
}
IOS Xcode project configuration for universal links
Open the Xcode project and go to the Info tab create a new URL type to be used for Dynamic Links.
Enter a unique value in Identifier field and set the URL scheme field to be your bundle identifier, which is the default URL scheme used by Dynamic Links.
In the Capabilities tab, enable Associated Domains and add the following to the Associated Domains list: applinks:your_dynamic_links_domain
(!) This should be only the domain - no https:// prefix
Android
Android doesn’t need additional configuration for default or custom domains.
Packages
A working react-native project setup with react-native-firebase is required, this is thoroughly covered in the library own documentation, here are the specific packages we used
Note: using the dynamicLinks package can be replaced with react-native's own Linking module and the code would be almost identical
Exact packages used:
"#react-native-firebase/app": "^6.7.1",
"#react-native-firebase/auth": "^6.7.1",
"#react-native-firebase/dynamic-links": "^6.7.1",
Sending the link to the user email
The module provides a sendSignInLinkToEmail method which accepts an email and action code configuration.
Firebase sends an email with a magic link to the provided email. Following the link has different behavior depending on the action code configuration.
The example below demonstrates how you could setup such a flow within your own application:
EmailLinkSignIn.jsx
import React, { useState } from 'react';
import { Alert, AsyncStorage, Button, TextInput, View } from 'react-native';
import auth from '#react-native-firebase/auth';
const EmailLinkSignIn = () => {
const [email, setEmail] = useState('');
return (
<View>
<TextInput value={email} onChangeText={text => setEmail(text)} />
<Button title="Send login link" onPress={() => sendSignInLink(email)} />
</View>
);
};
const BUNDLE_ID = 'com.example.ios';
const sendSignInLink = async (email) => {
const actionCodeSettings = {
handleCodeInApp: true,
// URL must be whitelisted in the Firebase Console.
url: 'https://www.example.com/magic-link',
iOS: {
bundleId: BUNDLE_ID,
},
android: {
packageName: BUNDLE_ID,
installApp: true,
minimumVersion: '12',
},
};
// Save the email for latter usage
await AsyncStorage.setItem('emailForSignIn', email);
await auth().sendSignInLinkToEmail(email, actionCodeSettings);
Alert.alert(`Login link sent to ${email}`);
/* You can also show a prompt to open the user's mailbox using 'react-native-email-link'
* await openInbox({ title: `Login link sent to ${email}`, message: 'Open my mailbox' }); */
};
export default EmailLinkSignIn;
We're setting handleCodeInApp to true since we want the link from the email to open our app and be handled there. How to configure and handle this is described in the next section.
The url parameter in this case is a fallback in case the link is opened from a desktop or another device that does not
have the app installed - they will be redirected to the provided url and it is a required parameter. It's also required to
have that url's domain whitelisted from Firebase console - Authentication -> Sign in method
You can find more details on the supported options here: ActionCodeSettings
Handling the link inside the app
Native projects needs to be configured so that the app can be launched by an universal link as described
above
You can use the built in Linking API from react-native or the dynamicLinks #react-native-firebase/dynamic-links to intercept and handle the link inside your app
EmailLinkHandler.jsx
import React, { useState, useEffect } from 'react';
import { ActivityIndicator, AsyncStorage, StyleSheet, Text, View } from 'react-native';
import auth from '#react-native-firebase/auth';
import dynamicLinks from '#react-native-firebase/dynamic-links';
const EmailLinkHandler = () => {
const { loading, error } = useEmailLinkEffect();
// Show an overlay with a loading indicator while the email link is processed
if (loading || error) {
return (
<View style={styles.container}>
{Boolean(error) && <Text>{error.message}</Text>}
{loading && <ActivityIndicator />}
</View>
);
}
// Hide otherwise. Or show some content if you are using this as a separate screen
return null;
};
const useEmailLinkEffect = () => {
const [loading, setLoading] = useState(false);
const [error, setError] = useState(null);
useEffect(() => {
const handleDynamicLink = async (link) => {
// Check and handle if the link is a email login link
if (auth().isSignInWithEmailLink(link.url)) {
setLoading(true);
try {
// use the email we saved earlier
const email = await AsyncStorage.getItem('emailForSignIn');
await auth().signInWithEmailLink(email, link.url);
/* You can now navigate to your initial authenticated screen
You can also parse the `link.url` and use the `continueurl` param to go to another screen
The `continueurl` would be the `url` passed to the action code settings */
}
catch (e) {
setError(e);
}
finally {
setLoading(false);
}
}
};
const unsubscribe = dynamicLinks().onLink(handleDynamicLink);
/* When the app is not running and is launched by a magic link the `onLink`
method won't fire, we can handle the app being launched by a magic link like this */
dynamicLinks().getInitialLink()
.then(link => link && handleDynamicLink(link));
// When the component is unmounted, remove the listener
return () => unsubscribe();
}, []);
return { error, loading };
};
const styles = StyleSheet.create({
container: {
...StyleSheet.absoluteFill,
backgroundColor: 'rgba(250,250,250,0.33)',
justifyContent: 'center',
alignItems: 'center',
},
});
const App = () => (
<View>
<EmailLinkHandler />
<AppScreens />
</View>
);
You can use the component in the root of your app as in this example
Or you can use it as a separate screen/route - in that case the user should be redirected to it after
the sendSignInLinkToEmail action
Upon successful sign-in, any onAuthStateChanged listeners will trigger with the new authentication state of the user. The result from the signInWithEmailLink can also be used to retrieve information about the user that signed in
Testing the email login link in the simulator
Have the app installed on the running simulator
Go through the flow that will send the magic link to the email
Go to your inbox and copy the link address
Open a terminal and paste the following code
xcrun simctl openurl booted {paste_the_link_here}
This will start the app if it’s not running
It will trigger the onLink hook (if you have a listener for it like above)
References
Deep Linking In React Native Using Firebase Dynamic Links
React Native Firebase - Dynamic Links
React Native Firebase - auth - signInWithEmailLink
firebase.google.com - Passing State In Email Actions
firebase.google.com - Authenticate with Firebase Using Email Link in JavaScript

how to use react-native-web and react-navigation together and access from web url

UPDATE:
react-navigation web support is done. follow this:
https://reactnavigation.org/docs/en/web-support.html
ORIGIN:
I try to share my code between react-native and web.
when I try react-native-web, it works well.
but there is only one question, how to access the specific screen from URL?
I read the react-navigation docs, there nothing about that.
and react-router-native can catch the web URL,
but it has navigator likes StackNavigator/DrawerNavigator.
and idea about that?
I'm not sure what the case was at the time you posted this question, but you definitely can use react-navigation with web now adays.
Now with Linking we can Handle deep links in React Native apps on Android and iOS, plus
Enable URL integration in browser when using on web.
The NavigationContainer component takes in a linking prop which allows you to map out your routes.
const linking = {
prefixes: ['https://mychat.com', 'mychat://'],
config: {
screens: {
Chat: 'feed/:sort',
Profile: 'user',
},
},
};
function App() {
return (
<NavigationContainer linking={linking} fallback={<Text>Loading...</Text>}>
{/* content */}
</NavigationContainer>
);
}
Once we establish what all of the routes or "links" are in our app we can start using Link components to navigate just like in a normal react web application if you used react-router-dom.
import { Link } from '#react-navigation/native';
// ...
function Home() {
return <Link to="/profile/jane">Go to Jane's profile</Link>;
}
These link components should work on both mobile, and web versions.
https://reactnavigation.org/docs/configuring-links/
https://reactnavigation.org/docs/deep-linking/
https://reactnavigation.org/docs/link
I don't think it's possible as ReactNavigation is using an internal state object. Remember, it's mobile framework, it has no concept of URL routing.
I also wanted to point out that even though RN claims web support you will need to be careful with component selection as not all the behaviours are identical (from memory, FlatList does not support touch scroll)