How to open fb and instagram app by tapping on button in Swift

Question

How can I open Facebook and Instagram app by tapping on a button in swift? Some apps redirect to the Facebook app and open a specific page. How can I do the same thing?

I found it:

var url = NSURL(string: "itms://itunes.apple.com/de/app/x-gift/id839686104?mt=8&uo=4")

if UIApplication.sharedApplication().canOpenURL(url!) {
  UIApplication.sharedApplication().openURL(url!)
}

but I have to know the app URL. Other examples were in ObjectiveC, which I don't know =/

Answer 1

Update for Swift 4 and iOS 10+

OK, there are two easy steps to achieve this in Swift 3:

First, you have to modify Info.plist to list instagram and facebook with LSApplicationQueriesSchemes. Simply open Info.plist as a Source Code, and paste this:

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

After that, you can open instagram and facebook apps by using instagram:// and fb://. Here is a complete code for instagram and you can do the same for facebook, you can link this code to any button you have as an Action:

@IBAction func InstagramAction() {

    let Username =  "instagram" // Your Instagram Username here
    let appURL = URL(string: "instagram://user?username=\(Username)")!
    let application = UIApplication.shared

    if application.canOpenURL(appURL) {
        application.open(appURL)
    } else {
        // if Instagram app is not installed, open URL inside Safari
        let webURL = URL(string: "https://instagram.com/\(Username)")!
        application.open(webURL)
    }

}

For facebook, you can use this code:

let appURL = URL(string: "fb://profile/\(Username)")!