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

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 =/

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)")!

Take a look at these links, it can help you:

https://instagram.com/developer/mobile-sharing/iphone-hooks/

http://wiki.akosma.com/IPhone_URL_Schemes

Open a facebook link by native Facebook app on iOS

Otherwise, there is a quick example with Instagram for opening a specific profile (nickname: johndoe) here:

var instagramHooks = "instagram://user?username=johndoe"
var instagramUrl = NSURL(string: instagramHooks)
if UIApplication.sharedApplication().canOpenURL(instagramUrl!) {  
  UIApplication.sharedApplication().openURL(instagramUrl!)
} else {
  //redirect to safari because the user doesn't have Instagram
  UIApplication.sharedApplication().openURL(NSURL(string: "http://instagram.com/")!)
}

You actually don't need to use a web and app URL anymore. The web URL will automatically open in the app if the user has it. Instagram or other apps implement this on their end as a Universal Link

Swift 4

func openInstagram(instagramHandle: String) {
    guard let url = URL(string: "https://instagram.com/\(instagramHandle)")  else { return }
    if UIApplication.shared.canOpenURL(url) {
        if #available(iOS 10.0, *) {
            UIApplication.shared.open(url, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(url)
        }
    }
}

In swift 3;

First you should add this on your Info.plist

Than you can use this code;

    let instagramUrl = URL(string: "instagram://app")
    UIApplication.shared.canOpenURL(instagramUrl!)
    UIApplication.shared.open(instagramUrl!, options: [:], completionHandler: nil)