using some basic code, I am trying to access a sharepoint page:
import { WebView } from 'react-native-webview'
const ExampleScreen = ({ navigation }) => {
return (
<WebView style={{flex: 1}}
source={{uri: 'https://example.intra.net/sites/1/ProjectName/SitePages/Nameofpage.aspx'}}/>
)
}
export default ExampleScreen;
When I pull up the page on Safari, I can see it, because I'm logged in. However when I try to look at the page in the Expo Go test build, it says "401 Unauthorized". What is the process for authenticating something like this in a native app? I am a novice and I don't understand how it works.
Related
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
I am learning RN, and as a part of my project, I am building an app that will take me to my toolbelt of items I will be making or have made such as animations. I am currently struggling with the login component of it. I am looking to login to my app so that it can take me to the homepage. I am using asyncstorage. When I log in, nothing happens except it stays on the login page. I have provided the lines of code as to how I have it built out. I really am just looking for something to get to the homepage on my simulator. I won't be deploying it or anything as I am keeping it to myself, but if any of you have any recommendations for articles or videos on how to achieve this in a "simpler" fashion, I am open to that option as well.
<TouchableOpacity
style={styles.buttonContainer}
onPress={this._signInAsync}
>
<Text style={styles.buttonText}>LOGIN</Text>
</TouchableOpacity>
</View>
);
}
_signInAsync = async () => {
await AsyncStorage.setItem("userToken", "abc");
this.props.navigation.navigate("app");
};
}
I am looking for an easy way to open apps, Facebook and Instagram specifically, on a button press from my React Native app. It should also check if the app is installed on the device first, and open the app store if it isn't. It needs to work on both iOS and Android. I am a beginner so if you can post an example it would help.
You can use Linking module of react-native to open other mobile apps.
import { Linking } from "react-native";
const APP_ID = //ID of app need to open in play store
const appDeepLinkURL = //Most of the mobile app provide it
Linking.openURL(appDeepLinkURL).catch(err => {
Linking.openURL(
`market://details?id=${APP_ID}`
).catch(err => Linking.openURL(
`http://play.google.com/store/apps/details?id=${APP_ID}`
).catch(err => console.error("An error occurred", err)););
});
Similarly, you can do for the iOS,
you can refer to the official doc here.
use react-native Linking Component
import {
TouchableOpacity,
Text,
Linking,
} from 'react-native';
<TouchableOpacity onPress={() => { Linking.openURL('sms:' + {contactNumber}
+ '?body=Hi'); }}>
<Text> Open Message App </Text>
</TouchableOpacity>
I've been trying to build a react native app that requires users to authenticate with their Instagram account. The Instagram API has a authorisation link and perhaps the only way to display that in an app would be through 'WebView' and so I used that.
The authentication workflow runs smoothly and then my server even gets the access token and user-id. But the problem is how to send this access token back to the app? I've used express-js for the 'redirect-uri' and so the WebView makes request to app.get() handler. In order to send response to same client on which the connection is opened, we must use res.send(). This would send the response to WebView, let's say I capture that using 'injectedJavaScript' but this javascript runs within WebView and so its unable to access react-native variables. In the event of a correct access-token, how would I ever navigate away from the WebView?
Any solutions to the above problems would be greatly appreciated. I suspect that there might even be problems with this approach(in my choice of WebView for this purpose, etc.), so a change of approach even entirely would also be of help. All I want is to authenticate the app users with Instagram and get my project going. Thanks a lot.
If you are using Expo, you can use AuthSessions to accomplish this (https://docs.expo.io/versions/latest/sdk/auth-session). The exact way to do it depends on whether you are using a managed workflow or a bare workflow, etc., but for managed workflow you can do the following:
Go to the Facebook Developer's console, go to your app, and add the Instagram Basic Display product.
Under Instagram Basic Display, under Valid OAuth Redirect URIs, use https://auth.expo.io/#your-expo-username/your-project-slug (project slug is in your app.json file)
On the same FB Developer page, add an Instagram tester profile and then follow the steps to authenticate that user.
In your project, install expo install expo-auth-session and import it into your Component
Also install expo-web-browser
Code your component like so:
import React, { useEffect } from 'react';
import { Button, Platform, Text, TouchableOpacity, View } from 'react-native';
import * as WebBrowser from 'expo-web-browser';
import { useAuthRequest, makeRedirectUri } from 'expo-auth-session';
WebBrowser.maybeCompleteAuthSession(); // <-- will close web window after authentication
const useProxy = Platform.select({ web: false, default: true });
const client_id = 9999999999999;
const redirect_uri = "https://auth.expo.io/#your-expo-username/your-project-slug";
const scope = "user_profile,user_media";
const site = "https://api.instagram.com/oauth/authorize?client_id=" + client_id + "&redirect_uri=" + redirect_uri + "&scope=" + scope + "&response_type=code&state=1";
const discovery = { authorizationEndpoint: site }
const GetInstagram = () => {
const [request, response, promptAsync] = useAuthRequest({
redirectUri: makeRedirectUri({
useProxy,
native: redirect_uri
}),
scopes: [scope],
clientId: client_id
}, discovery);
useEffect(() => {
if (response?.type === 'success') {
const { code } = response.params; <--- the IG code will be returned here
console.log("code : ", code);
}
}, [response]);
return (
<View>
<TouchableOpacity onPress={ () => promptAsync({useProxy,windowFeatures: { width: 700, height: 600 }}) }>
<Text>Connect Your Instagram</Text>
</TouchableOpacity>
</View>
)
}
export default GetInstagram;
One way to accomplish this is via using deeplink. I don't think it's the best practice though.
The response from the WebView will be sent to the redirect URL you've previously setup after successful authentication. Please set the redirect URL to your app. For example, on iOS if you have URL Scheme as "myapp123" then, anytime you open your browser and type myapp123://.. it will open your app and your app should be able to get response sent from the instagram.
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)