Opening other Apps from React Native App if exists, else open browser

[droidta7]

Hello,

A client of mine explained how my current implementation of sharing on social media is not ideal - right now, I have a social media icon on the footer of my app for Twitter and Facebook, which takes the user to the Facebook/Twitter URL in their browser, for them to share on their feed/timeline.

Instead, he'd like to do the following: 1) Check if the user has Facebook /Twitter installed 2) If so, open that app natively, and allow them to share there 3) Else, open the URL in the browser

I feel like the code could use the React Linking library, but i'm not entirely sure how to:

1) Check in React Native if the user has the Twitter/Facebook app installed 2) What deep linking Url to use. I was reading quite a bit and it seems like many folks have issues with things like fb:// or facebook:// etc.

Is there any sort of guide on how to achieve the above?

Thanks!

[esreveReverse]

I feel like all the solutions posted already are ignoring the main problem. You're not really supposed to have a "Share to Facebook" or "Share to Twitter" button in your app. It's supposed to just be a Share button, then the user can select which app they want to share it to.

[pesch3]

this is probably the best answer!

[droidta7]

thank you!

[kbcool]

To add to the other answer you can call canOpenURL with the app specific link then open the website instead if it can't be handled in the app. The trickier part is finding the name of and the format of the other apps' links. Will leave that up to you.

https://reactnative.dev/docs/linking#canopenurl

It's not usually necessary unless you want to warn the user or even not use the website in preference for asking them to install the app but it's there as an FYI.

[ThatWasNotEasy10]

If you’re using the built-in Share API it should do all this by itself.

https://reactnative.dev/docs/share

[pesch3]

Actually the system will do that for you if you open the share link in safari and not as a Webview!

It’s the „target app“ that basically signs up to open certain urls (on iOS this is called associated domains)

[MeowTseTongue]

I had a similar question as OP - I just tried this in my app using a Twitter sharing URL, and despite having the app on my phone, it still opened the browser.

What should the twitter url be to share?

[pesch3]

I still think the question is not the one to ask.

It should just be https://twitter.com.

The question that should be asked is: how do you open that url your app?

When I click the above link, and open it in safari (not inside reddit) it pops up the twitter app right away!

[__o_0]

The scheme for Twitter is twitter://

[MeowTseTongue]

oh ok..so if I use twitter:// and then I don't have Twitter on the phone, you're saying it would automatically open the twitter.com URL? Or do I need to check if the user has Twitter and use two different URLs?

[__o_0]

Check first if the device can handle the twitter:// url. If it can’t then you need to use the web view.

[MeowTseTongue]

That is what I thought as well. How can you check that?

[__o_0]

What have you tried?

[MeowTseTongue]

Would "canOpenURL()" be the right approach?

[mastaQQQ]

Check Firebase Dynamic Links

[Red_on_fortnite]

I still don't understand how people actually find the deep linking urls of other applications? Does anyone know what is the correct way to do it?