React Native open Facebook page

I'm having the same issue, but not able to solve it. I have multiple social networking buttons with the onpress action of onPress= {()=> Linking.openURL("https://www.SOME_SOCIAL.NETWORK")} My buttons linking to Twitter, Instagram, and SnapChat all open the app if installed or a web page in Safari if the app is not installed. The only outlier is Facebook. Given an onpress action such as onPress= {()=> Linking.openURL("https://www.facebook.com/")} the link will always open in Safari even if the app is installed.

Because of this odd behavior I'm handling my onpress action for the Facebook link like this:

```

  Linking.canOpenURL("fb://profile/XXXXXX").then(supported => {
    if (supported) {
      return Linking.openURL("fb://profile/XXXXXX");
    } else {
      return Linking.openURL("https://www.facebook.com/");
    }
  })

```

The above code doesn't work as intended, but nothing else seems to.

Update - May 2019

// open via app

fb://facewebmodal/f?href=PAGE_NAME

In general, you should check Linking.canOpenURL() for iOS before trying to open them.

Also, make sure you add your protocols as an array in info.plist as:

<key>LSApplicationQueriesSchemes</key>
<array>
  <string>fb</string>
</array>

You can also do this in your Xcode project in info.plist.

Ok, found one way to open Facebook:

Linking.openURL('fb://page/PAGE_ID');
Linking.openURL('http://instagram.com/_u/USER_NAME');
Linking.openURL('http://instagram.com/_p/PICTURE');