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

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)        }    }}