Showing and Popping Multiple View Controllers - ios

So here's my problem.
First ViewController has some fields regarding name, date, no. of people of an Event (the user creates and event). It also has a button that will transition the user to a new viewController with a mapView where user will create a Pin. After user creates a Pin, user can press the pin (or do some other action that confirms that the Pin user put on the map is correct) and user will be transitioned to a new tableViewController with multiple locations near the Pin he selected (using Foursquare API).
Then the user must be transitioned to the first ViewController. The Location user tapped must be sent to the first ViewController.
This is my first time working with multiple ViewControllers and I need your help.
How would I do this (don't miss on any details please :D) ?

Use an unwind segue to pass data backwards through view controllers embedded in a navigation controller.
Here's a brief example.
In your first view controller you can have something like:
var text: String?
#IBAction func unwindSegue(segue: UIStoryboardSegue) {
if let viewController = segue.sourceViewController as? ViewController2 {
text = viewController.textField.text
}
print(text!)
}
Add the unwind segue in the second view controller on the storyboard by ctrl-dragging from the view controller (yellow circle with a square in, to the exit icon (either at the top of the view controller or in the document outline) and click on the unwind segue that you created to create a manual segue. Then in the document outline set the unwind segue's identifier so that you can refer to it (I've called it simply "Unwind" for this example).
In the second view controller (the one you want to pass back from) add this:
#IBOutlet var textField: UITextField!
#IBAction func backButton(sender: AnyObject) {
self.performSegueWithIdentifier("Unwind", sender: nil)
}
This simple example will let you pass back the text from the textField in your second view controller.
n.b. You can unwind more than one view at a time.

Related

How to segue from tab bar view controller to another view which is a child of a navigation controller

I laid out my home screen (Summary screen) as shown on the screenshot below:
I added the "+" button on the UITabBarController subclass. This button will segue to a view controller and show it modally. To give you a bit of context on how I've structured my storyboards, please refer to the screenshot below:
This modal view controller has a form that the user needs to fill in, and once they're done, there's a 'Done' button which will 1) dismiss the modal, 2) take them back to the root view which is either the 'Summary' tab or the 'Details' tab then 3) take them to a list view showing the most recent data they've entered on a list (destination view – highlighted in red).
Now, doing the segue when the user taps the "+" button is simple enough. With the following code:"
menuButton.addTarget(self, action: #selector(menuButtonAction(sender:)), for: .touchUpInside)
then
func addNewExpense(action: UIAlertAction){
self.performSegue(withIdentifier: "segue_formModal", sender: self)
}
On the modal I've setup a protocol which will then send the form data back to the tab view controller.
extension XtabBarViewController: NewInfoDelegate{
func newInfoSubmitted(formData: FormData) {
performSegue(withIdentifier: "segue_ListingPage", sender: self)
}
}
But it seems that the perform segue is not working. It says the segue does not exist. "segue_ListingPage" is that segue that connects the Summary view and the destination view.
How can I segue from the tab bar view controller to the destination view? Any help is appreciated.
Thanks in advance!
You can just make the navigation view controller the entry point.
I figured it out on my own. For the benefit of those who run into the same issue here's my solution.
I simply added the code from the tab bar view controller the code:
let childVC = children[0].children[0]
childVC.performSegue(withIdentifier: "segue_TXListPage", sender: self)
So what I did here is call the performSegue from the view controller that segues to the destination controller.

Pass data between three viewController, all in navigationController, popToRootView

The issue I'm having is this.
I have a navigation controller with 3 viewController. In the 1st controller, I have the user select an image. This image is passed to 2nd and 3rd controller via prepareForSegue.
At the 3rd controller, I have a button that takes the user back to the 1st view controller. I explored 2 ways in doing this:
1) use performSegue, but I don't like this because it just push the 1st controller to my navigation stack. So I have this weird "Back" button at the 1st Viewcontroller now, which is not what I want. I want the app to take user directly to 1st viewcontroller without the back button.
2) I tried Poptorootviewcontroller. This solves the issue of the "back" button. But, when I pop back to the 1st viewcontroller, the user's selected image is still on screen. I want to clear this image when the user goes from the 3rd viewcontroller back to the 1st viewcontroller.
So with approach 2), how do I make sure all memory is refreshed and the image becomes nil in the 1st viewcontroller? Since I'm not using performSegue, 3rd viewcontroller does not have access to the 1st Viewcontroller.
For refresh, you'd have to clear it in viewWillAppear but I find this rather dangerous. Best you can do there is to create a new copy of the view controller everytime and Swift will take care of the rest. I don't know if you are using the storyboard but I would recommend using the class UIStoryboard and the function instiantiateViewControllerWithIdentifier("something") as! YourCustomVC
As long as you stay in the navigation stack, you'll not lose any of the current configurations of previous View Controllers.
As for passing data back to the first controller. You can either just throw it in the global scope which is the easiest way but might be difficult to know when it was updated or if the data is fresh. But you can always just:
var something: String = ""
class someView: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
something = "foo"
}
}
Something will be availabe everywhere then.
You could make a protocol and pass the delegate along the 3 view controllers. So when you are starting it you could do:
func someAction() {
let v = SomeViewController()
v.delegate = self
self.navigationController?.pushViewController(v, animated: true)
}
And then with each following view:
func someOtherAction() {
let v = SomeOtherViewController()
v.delegate = self.delegate
self.navigationController?.pushViewController(v, animated: true)
}
Although personally I find it hard to keep track of this.
Lastly you could use the NSNotificationCenter to pass an object along with all the data and catch it in a function on your first controller.
To do this you first register your VC for the action in viewDidLoad() or something:
NSNotificationCenter.defaultCenter().addObserver(self, selector: "someAction:", name: "someNotification", object: nil)
Then when you are done in the 3rd view make some object or a collection of string and send it back as follows:
NSNotificationCenter.defaultCenter().postNotificationName("someNotification", object: CustomObject())
And then lastly you'll catch it in the function "someAction"
func someAction(note: NSNotification) {
if let object = note.object as? CustomObject {
//Do something with it
}
}
Hope this helps!
Use an unwind segue which provides the functionality to unwind from the 3rd to the 1st (root) view controller.
The unwind segue is tied to an action in the root view controller. Within this action, you simply nil the image:
#IBAction func unwindToRootViewController(sender: UIStoryboardSegue)
{
let sourceViewController = sender.sourceViewController
// Pull any data from the view controller which initiated the unwind segue.
// Nil the selected image
myImageView.image = nil
}
As you can see in the action, segues also let you pass data back from the source view controller. This is a much simpler approach than needing to resort to using delegates, notifications, or global variables.
It also helps keep things encapsulated, as the third view controller should never need to know specifics about a parent view controller, or try to nil any image that belongs to another view controller.
In general, you pass details to a controller, which then acts on it itself, instead of trying to manipulate another controller's internals.

Multiple unwind in the same button - Swift

I would like to know if it is possible to assign two different unwind method at the same button. For example:
I have this views navigations:
Home->A->B
Home->C->B
A and C views navigates to B, but then I want to return to previous views using the same B view and controller.
It is possible?
I have been thinking about write assign unwind method to the button programmatically depending what view comes.
Thanks in advance
I'm sorry about my english, is not good.
Here's a Swift solution that worked well for me. The code below only works if you hookup your segues correctly in the storyboard and in code. Checkout this page for great explanations on setting up unwind segues.
In summary:
You're accessing the same view from multiple other views. So, when you segue to a view, you can pass the source view controller (the view that you're currently in) to a property in the view that you're going to.
In your view that you will unwind out of, you can check the property holding the info (the class) on where you came from, and then perform a segue based on what view it is.
The code: (using ex: Home -> A -> B or... Home -> C -> B)
Note: B is the view that will unwind to multiple different views.
In A or C: (code works the same way in both views)
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if segue.identifier == "segueIdentifierInViewAthatGoesToViewB" {
let controller:B = segue.destinationViewController as! B
//the code below passes our current view controller class to a property in view B.
//So, view B will know what view we came from.
//In our example, we're coming from view A or C
controller.viewControllerNavigatedFrom = segue.sourceViewController
}
}
In B:
//setup an IBAction that will check the source view controller (that we passed when we segued to this view) and perform a segue based on where we came from. You can hook this up to a button or anything you want really.
//isKindOfClass(A) - "A" would be the name of your class
//setup a property to receive the view controller class where we are coming from
var viewControllerNavigatedFrom:AnyObject?
#IBAction func myButtonPressed(sender: AnyObject) {
if self.viewControllerNavigatedFrom!.isKindOfClass(A) {
//Unwind to view A
performSegueWithIdentifier("unwindFromBbackToA", sender: sender)
}
else if self.viewControllerNavigatedFrom!.isKindOfClass(C) {
//Unwind to view C
performSegueWithIdentifier("unwindFromBbackToC", sender: sender)
}
}
Although, question isn't very clear. But what I could understand is that you want to navigate back to the previous view i.e. B>C or B>A depending upon where user came from.
If so, then check the UINavigationController. It keeps track of the navigation history and automatically adds a back button. Kind of like the back button in our browsers.
Here is a tutorial, although a bit old: Link

Right way or event to choose what view load in swift

I'm working in an app that logs in an user if there isn't another user already logged in at launch time. This way the first view to appear should be the Login View. But in the case there is a logged user already, the first view appearing should be the main menu. Im handling this with the viewWillAppear function and it's working, but I don't know if this is the correct approach or how it should be handle in this situations.
Here is my code. My first view is MainMenuVC in which I control if there is a logged user or not, then I choose if stay in main menu view or push my login view.
class MainMenuVC: UIViewController {
override func viewWillAppear(animated: Bool) {
if (UserMgr.users.count == 0){
var vc1:LoginVC = self.storyboard?.instantiateViewControllerWithIdentifier("LoginView") as LoginVC
self.navigationController?.pushViewController(vc1, animated: false)
}
else
{
//I do nothing so this view is loaded
}
}
I don't know if i should use another ViewController and implement the function loadView() to decide what view load, but the problem is make that view work with the story board and my navigation controller.
Any suggestions?
Basically you will have two different view controllers, one for the login screen (VCLogin) and one for the main menu (VCMainMenu). Now, in your AppDelegate there are methods which are called, when the app launches respectively when it appears. So, place the code checking whether a user is logged in there and make the appropriate view controller the root view controller, e.g.
let navigationController = window.rootViewController as UINavigationController
navigationController.rootViewController =
userIsLoggedIn ? mainMenuViewController : loginViewController

Swift: Perform a function on ViewController after dismissing modal

A user is in a view controller which calls a modal. When self.dismissViewController is called on the modal, a function needs to be run on the initial view controller. This function also requires a variable passed from the modal.
This modal can be displayed from a number of view controllers, so the function cannot be directly called in a viewDidDisappear on the modal view.
How can this be accomplished in swift?
How about delegate?
Or you can make a ViewController like this:
typealias Action = (x: AnyObject) -> () // replace AnyObject to what you need
class ViewController: UIViewController {
func modalAction() -> Action {
return { [unowned self] x in
// the x is what you want to passed by the modal viewcontroller
// now you got it
}
}
}
And in modal:
class ModalViewController: UIViewController {
var callbackAction: Action?
override func viewDidDisappear(_ animated: Bool) {
let x = … // the x is what you pass to ViewController
callbackAction?(x)
}
}
Of course, when you show ModalViewController need to set callbackAction like this modal.callbackAction = modalAction() in ViewController
The answer supplied and chosen by the question asker (Michael Voccola) didn't work for me, so I wanted to supply another answer option. His answer didn't work for me because viewDidAppear does not appear to run when I dismiss the modal view.
I have a table and a modal VC that appears and takes some table input. I had no trouble sending the initial VC the modal's new variable info. However, I was having trouble getting the table to automatically run a tableView.reloadData function upon dismissing the modal view.
The answer that worked for me was in the comments above:
You likely want to do this using an unwind segue on the modal, that
way you can set up a function on the parent that gets called when it
unwinds. stackoverflow.com/questions/12561735/… – porglezomp Dec 15
'14 at 3:41
And if you're only unwinding one step (VC2 to VC1), you only need a snippet of the given answer:
Step 1: Insert method in VC1 code
When you perform an unwind segue, you need to specify an action, which
is an action method of the view controller you want to unwind to:
#IBAction func unwindToThisViewController(segue: UIStoryboardSegue) {
//Insert function to be run upon dismiss of VC2
}
Step 2: In storyboard, in the presented VC2, drag from the button to the exit icon and select "unwindToThisViewController"
After the action method has been added, you can define the unwind
segue in the storyboard by control-dragging to the Exit icon.
And that's it. Those two steps worked for me. Now when my modal view is dismissed, my table updates. Just figured I'd add this, in case anyone else's issue wasn't solved by the chosen answer.
I was able to achieve the desired result by setting a Global Variable as a boolean value from the modal view controller. The variable is initiated and made available from a struct in a separate class.
When the modal is dismissed, the viewDidAppear method on the initial view controller responds accordingly to the value of the global variable and, if needed, flips the value on the global variable.
I am not sure if this is the most efficient way from a performance perspective, but it works perfectly in my scenario.

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