Facebook Login button for Swift
Solution 1
One option would be to set your view controller as a delegate of the FBSDKLoginButton
and implement the loginButton:didCompleteWithResult:error:
method, which is called when the button is used to login.
Swift
class ViewController: UIViewController, FBSDKLoginButtonDelegate {
@IBOutlet weak var loginButton: FBSDKLoginButton!
override func viewDidLoad() {
super.viewDidLoad()
self.loginButton.delegate = self
}
}
Obj-C
// ViewController.h
@interface ViewController : UIViewController <FBSDKLoginButtonDelegate>
@property (weak, nonatomic) IBOutlet FBSDKLoginButton *loginButton;
@end
// ViewController.m
@implementation ViewController
- (void)viewDidLoad {
[super viewDidLoad];
self.loginButton.delegate = self;
}
Then, in the loginButton:didCompleteWithResult:error:
method you can check the result
and error
, and if everything is fine, navigate to another view.
Swift
func loginButton(loginButton: FBSDKLoginButton!, didCompleteWithResult result: FBSDKLoginManagerLoginResult!, error: NSError!) {
if ((error) != nil) {
// Process error
}
else if result.isCancelled {
// Handle cancellations
}
else {
// Navigate to other view
}
}
Obj-C
// ViewController.m
@implementation ViewController
- (void)loginButton:(FBSDKLoginButton *)loginButton
didCompleteWithResult:(FBSDKLoginManagerLoginResult *)result
error:(NSError *)error {
if (error) {
// Process error
}
else if (result.isCancelled) {
// Handle cancellations
}
else {
// Navigate to other view
}
}
You can find more about how to login with FB in their docs.
Solution 2
In Swift that would be something like:
class MyViewController: UIViewController, FBSDKLoginButtonDelegate {
@IBOutlet weak var loginView : FBSDKLoginButton!
@IBOutlet weak var profilePictureView : FBSDKProfilePictureView!
override func viewDidLoad() {
super.viewDidLoad()
self.loginView.delegate = self
if (FBSDKAccessToken.currentAccessToken() != nil)
{
performSegueWithIdentifier("unwindToViewOtherController", sender: self)
}
else
{
loginView.readPermissions = ["public_profile", "email", "user_friends"]
}
}
func loginButton(loginButton: FBSDKLoginButton!, didCompleteWithResult result: FBSDKLoginManagerLoginResult!, error: NSError!) {
println("User Logged In")
if ((error) != nil)
{
// Process error
}
else if result.isCancelled {
// Handle cancellations
}
else {
// If you ask for multiple permissions at once, you
// should check if specific permissions missing
if result.grantedPermissions.contains("email")
{
// Do work
}
}
}
func loginButtonDidLogOut(loginButton: FBSDKLoginButton!) {
println("User Logged Out")
}
}
Then in your TargetViewController add an unwind function:
@IBAction func unwindToViewOtherController(segue:UIStoryboardSegue) {
}
Solution 3
In current FacebookLogin version (0.2.0) for Swift, the LoginButton delegate property is defined as a strong property:
public class LoginButton: UIView {
...
/// Delegate of the login button that can handle the result, logout events.
public var delegate: LoginButtonDelegate?
... }
If you add the login button following Facebook instructions and you set your UIViewController
child class as button delegate...
import FacebookLogin
func viewDidLoad() {
let loginButton = LoginButton(readPermissions: [ .PublicProfile ])
loginButton.center = view.center
loginButton.delegate = self
view.addSubview(loginButton)
}
... a reference cycle will be created. The view will contain a strong reference to the button, the button will contain a strong reference to the controller, and the controller will have a strong reference to its view, see this post.
My solution was to use a weak member variable to have a reference to the login button and when the view disappears, the button delegate is set to nil, like this:
import UIKit
import FacebookCore
import FacebookLogin
import RxSwift
class LoginViewController: UIViewController, LoginButtonDelegate {
private weak var facebookLoginButton: LoginButton? = nil
override func viewDidLoad() {
super.viewDidLoad()
// Add the Facebook login button
let loginButton = LoginButton(readPermissions: [ .publicProfile, .email, .userFriends ])
loginButton.center = view.center
// WARNING!: Facebook login button delegate property is defined currently as STRONG.
// Therefore, it must be set to nil before leaving the view to avoid reference cycles
loginButton.delegate = self
view.addSubview(loginButton)
// Store the login button as a weak reference, since it is holded by the main view with a
// strong reference
facebookLoginButton = loginButton
}
override func willMove(toParentViewController parent: UIViewController?) {
super.willMove(toParentViewController:parent)
if parent == nil {
// The back button was pressed, interactive gesture used, or programatically pop view
// was executed
// Do not forget to set delegate in Facebook button to nil to break reference cycle.
facebookLoginButton?.delegate = nil
}
}
// MARK: - Facebook login
/**
Called when the button was used to login and the process finished.
- parameter loginButton: Button that was used to login.
- parameter result: The result of the login.
*/
func loginButtonDidCompleteLogin(_ loginButton: LoginButton, result: LoginResult) {
switch result {
case .failed(let error):
// Action on failed
case .cancelled:
// Action on cancelled
case .success(let grantedPermissions, let declinedPermissions, let accessToken):
// Action on success
}
}
/**
Called when the button was used to logout.
- parameter loginButton: Button that was used to logout.
*/
func loginButtonDidLogOut(_ loginButton: LoginButton) {
// Action on logout
}
}
Do not use function viewWillDissapear()
for setting to nil
the delegate, because Facebook login page will be shown on top of your app, triggering this function, and you will not get the login result since you will not be the delegate anymore. Note that this solution is working fine for views inside a navigation controller. Another solution should be found for modal windows.
I hope it helps, Xavi
Solution 4
IOS 13 use Scene Delegate. Just paste the below code in scene delegate and simple call the facebook login manager it will return the user object of facebook. This function automatically call.
func scene(_ scene: UIScene, openURLContexts URLContexts: Set<UIOpenURLContext>) {
guard let url = URLContexts.first?.url else {
return
}
let _ = ApplicationDelegate.shared.application(
UIApplication.shared,
open: url,
sourceApplication: nil,
annotation: [UIApplication.OpenURLOptionsKey.annotation])
}
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Soporificdreamer
Updated on July 01, 2022Comments
-
Soporificdreamer almost 2 years
In Xcode if I create a
UIView
and then add the custom class asFBSDKLoginButton
, when I click it leads me through the Facebook login and then returns me to the same page as theFBSDKLoginButton
but instead of saying login button it says log out now. How would I go about making that when the login button is clicked it lead to a new view?
I downloaded the Facebook SDK through cocoapods and its my first time working with it so I am confused about this. Thanks for the help!
-
J Richard Snape about 9 yearsThis came to me in a review queue as it was marked as a Link only answer, which are generally not seen as good on here.. I can see you've responded to the comment about link only answers. I edited to tidy up a bit of formatting / typos.
-
Nate Uni almost 9 yearsThanks bud.. a snippet of that proved handy!
-
rAzOr almost 9 years
performSegueWithIdentifier("unwindToViewOtherController", sender: self)
insideif (FBSDKAccessToken.currentAccessToken() != nil)
isn't loading next viewcontroller. Would u please help me with this?? -
Ruud Kalis almost 9 yearsIn the storyboard select the unwind segue and then the attribute inspector. There you should have
unwindToViewOtherController
as Identifier andunwindToViewOtherController:
as Action. -
rAzOr almost 9 yearsI had to add
viewDidAppear
function and writeperformSegueWithIdentifier
in it. That solved the issue. We can't haveperformSegueWithIdentifier
insideviewDidLoad
function, that's what I learned from google. Would you please help me know why? -
Ruud Kalis almost 9 yearsYou cannot dismiss a view that has not loaded yet.
ViewDidLoad
it is not after this function is finished that the view is presented and thus can be dismissed. -
GhostCat over 6 yearsIf you drop the lines saying "this is a comment" ... this would actually look like an answer to me ;-)