I am building a react native application in which I have to hit a URL to let user for a video call ,but when I open that URL in Webview I cant access camera and microphone of device(both a\Android and IOS) required for call and I can't open the URL in browser using linking because I can't let user know that URL because if they know URL they can call anytime using that URL but I don't want that.
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I'm working on a Facebook video downloader application in react native with webview. I have added a download button with every video so when user click on the download button I'm extracting that video src URL value and that is a blob value which is not an exact video url.
Blob URL: blob:https://m.facebook.com/fa1b8a59-f983-4881-971b-47f9fed39242
So can anyone please explain how to get the actual video URL in webview to download that specific video?
Often when viewing a website on a mobile device you'll get a notification that a mobile app is available, or if you already have the app downloaded you'll have the option to open the URL in the app.
Is there a native way to do this or are these notifications always custom?
When users click on a link or download a file and the suggestion pops up to open the respective app, that's called deep linking. iOS labels it as Universal Links and Android uses App Links.
I've created a React-native app (tested in android) using WebView with the initial URL of my company's website.
I want the user to be able to get redirected to a spesific website's product page (for example https://www.test.com/product/hoodie-with-zipper/) WHEN they clicked the OneSignal push notification with the intended launch URL included in the sent notification. The problem is, I want the user to open the URL with my WebView App, not the browsers.
The features I've implemented and tested so far:
Implemented and Tested WebView (no problems so far)
Implemented and Tested deeplink locally (with android scheme of 'myapp://', no problems so far)
Implemented OneSignal push notification SDK for react-native and its firebase requirements (no problems so far)
As I've mentioned in the point 2 before, I've tried and successfully implemented the deeplink with the android scheme of 'myapp://'. But, when I tried to implement the android scheme of 'https://', the launch URL from OneSignal push notification got opened with browser instead, not my WebView app.
I want the user to be able to get redirected to a spesific website's product page (for example https://www.test.com/product/hoodie-with-zipper/) WHEN they clicked the OneSignal push notification with the intended launch URL included in the sent notification, but it'll be opened with React-native WebView instead of browser.
^ So, what should I do to achieve that?
Thanks in advances.
You should handle this yourself by passing in the URL via additional data in the notification. Then, in your app you should open a webview to the specified url.The Launch URL feature on the dashboard or the url field on the REST API automatically opens the web browser on the mobile device when the notification is tapped on. If you would like to open the URL inside your app instead of the browser, you need to send your notification with custom data that can read by your app's SDK. If you're using our dashboard to send notifications, the sending notification options allows you to include additional data to be sent with your notification. If you're using our API to send notifications, you can set the 'data' field to a JSON object hash of extra custom data. For example, you could set targetUrl for the key and https://google.com as the value. Then in your app's code read the targetUrl value from additionalData in the NotificationOpened callback (name depends on the OneSignal SDK used).
I use react-native-simple-auth in my project but i can login twitter in web.I want to login with twitter app instead of safari view in react native. How to do this? Any Idea?
If you have installed Twitter in the phone use this:
Linking.openURL('twitter://app')
This is the snippet that open an external app. It open but you can't pass params to log in. Only open the app and, if you are logged, you will automatically enter to the app logged, else, just will open Login view.
I want to integrate a website into a webview with react native.
Everything is fine until I want to login with facebook. When I press the "Login With Facebook" button, the webview become just a white screen and it is blocked.
This happens because when I press the login button, it tries to open a popup, but the popups are not enabled in react native webview.
So, my question is: how is possible to enable popups in webview? I just want my website to behave just like in a normal browser (like chrome, where everything is working fine), but inside my webview.
I use react native 0.55.
I read https://codeburst.io/webviews-and-social-authentication-with-react-native-cfecf96ac7d7 but didn't help me because I don't have access to the website source code.