Pop the current view using Segues/Storyboard on iOS 5 - ios

I am creating an app using iOS 5 SDK. I managed to push views using the Storyboard's Segues, but I cannot find the proper way to pop the current view and go back to the previous one.
I am not using any navigationController (the app doesn't have any top or bottom bars).
I don't think using modal or push segue the other way would be the solution as it instantiates a new controller.
Do I have to use a custom Segue with the opposite animation and deletion of the view at the end ? Or is there a better way ?

Storyboards in iOS 5 don't provide a "no-code" way to return from a segue -- that's something you'll need to implement yourself.
If you use "push" segues (which require a navigation controller), use the navigation controller's popViewControllerAnimated: method to undo the last push segue. (Or other methods to undo more; see the UINavigationController documentation.)
If you use "modal" segues, call dismissViewControllerAnimated:completion: on the view controller which presented the current view controller (which you can get from its presentingViewController property).
Update: In iOS 6 and later there's unwind segues for going "back" in a storyboard. It's still not a no-code solution -- and it shouldn't be, because you need to be able to do things like differentiating between "Done" and "Cancel" exits from a modal view controller. But it does let you put more of the semantic flow of your app into the storyboard. Apple has a tech note that describes them in detail, and they're also covered in the video from WWDC 2012 Session 407.

You could try calling [self dismissViewControllerAnimated:YES completion:nil]; from the controller you want to dismiss (whether the controller has been pushed, or shown modally).
Here is the related documentation : UIViewController Class Reference
The presenting view controller is responsible for dismissing the view controller it presented. If you call this method on the presented view controller itself, it automatically forwards the message to the presenting view controller.

Just to clarify.
In the class that was pushed. Simply wire up the following and the controller and view will be popped off.
[self.navigationController popViewControllerAnimated:YES];

Create Segue type "Custom" on your stroyboard. This can be from a button.
Create a new UIStoryboardSegue class named "popSegue"
In the popSegue.m file add the following;
-(void)perform{
UIViewController *sourceViewContreoller = [self sourceViewController];
[sourceViewContreoller.navigationController popViewControllerAnimated:YES];
}
-In the storyboard editor.
-Select the segue and change the Segue Class to "popSegue"
-Set the Identifier to "popSegue"
Done!
You can use the same "popSegue" class throughout your project.
Hope this helps

I'm using Xcode 5 also and here's how it's done. First, in the view code file that pushed the other, create an IBAction method in the .h file such as this:
- (IBAction)exitToHere:(UIStoryboardPopoverSegue *)segue sender:(id)sender;
Then in the .m file add this:
- (IBAction)exitToHere:(UIStoryboardPopoverSegue *)segue sender:(id)sender {
}
You can add any cleanup code you want executed in this method. Next go to your storyboard and select the pushed view. I assume you've got some kind of button on the view that the user taps to signal he's finished. Click on that button, hold down the key and drag to the the green box below the view which is the Exit. Release the mouse button but continue to hold the key. A popup will appear and your method will show in the list. Select that method. Now when the user clicks on the button, the view will pop and you'll be returned to the starting method.

Related

Segue from screen with several buttons?

I am trying to write an app using UINavigationViewController. My first screen has several buttons on it, and on the click of each button, I want to segue to a UIViewController. I know that I can add a segue on each button, all pointed to the UIViewController that I want to go to, but I was wondering if it is possible to use only one segue that can be fired from each of the buttons.
If that is not possible, I was wondering if it was possible to open the second UIViewController from the first one, on button click, and provide a Back button like the UINavigationView provides. I did manage to get everything on this idea working, except for the back button. I mean I can put a standard button somewhere on the screen and go back, but I'd like the standard back button on the UINavigationView.
Phew! I'm not sure if that makes any sense.
I know that I could also use a tableview, but I'm trying to set this up with buttons.
Thanks
Edit: Thank you to everyone that answered. I now have this working. I would vote up the answers, but I don't have enough posts to do it. I appreciate the answers!
If you need to have separate action functions for each button, suggest that you segue from the main controller to the other controller and create a segue identifier (see xcode procedure below); then, use performSegueWithIdentifier from each of the button action functions. You can also take advantage of the prepareForSegue. To create the segue, control-drag from the left button in the controller in the storyboard to the controller you want to segue to and pick show.
Check the example code in swift that I did for a very similar problem in the SO reference
Linking View Controllers through button
You can embed the main controller in a navigation controller and that will give you the ability to navigate back. If you have multiple layers you can also use unwind segue.
Link each button to one single action (ex. buttonClick) in that ViewController and then perform the appropriate segue using pushViewController method on self.navigationController
-(IBAction)buttonClick:(id)sender {
if(sender.id == self.button1) {
DestinationViewController *vc = [[UIStoryboard storyboardWithName:#"Main" bundle:[NSBundle mainBundle]] instantiateViewControllerWithIdentifier:#"VC_IDENTIFIER"];
[self.navigationController pushViewController:vc animated:YES];
}
Or if you already have that 1 segue defined in storyboards you can use
[self performSegueWithIdentifier:#"SegueIdentifier" sender:self];
And use that inside the buttonClick method. Using the 1st example, or the second one as long as the segue you setup in the storyboards is a push then you should already get the back button as that is the default behavior for pushing view controllers onto the navigation stack.

UIStoryboard Show (e.g. push) simple inverse action to go back

In storyboard we have great feature that allow us to make Show (e.g. push). So seems the logic is next:
If we don't have navigation controller then view controller will use present modal logic. My question is there any inverse action that I can use with Show?
I have a UIButton that close current view controller screen:
- (IBAction)onTappedCloseButton:(id)sender
{
[self.navigationController popViewControllerAnimated:YES];
}
But in case if I don't have navigation controller, how can I simple use inverse action to go back? So my solution is to check if self.navigationController is nil then use dismissing option:
[self dismissViewControllerAnimated:YES completion:nil];
But maybe there is another cool solution like Show (e.g push). But Close (e.g. pop)?
Yes, you can use an unwind segue to go back, and it will be the reverse of whatever the forward segue was.
You have two options on how to do this:
1) The Unwind segue
To make an unwind segue you have to add a method in the view controller you want to "unwind" to with the following format:
-(IBAction)someSelectorName:(UIStoryboardSegue *)sender;
You will then be able to drag from your UIButton up to the "exit" icon in your storyboard.
Wire it up to the selector you just defined and UIKit will figure out how to get back to that view controller without you having to write any code. This can be especially useful as it can figure out when it needs to call -dismissViewControllerAnimated: more than once and can call those methods successfully. It can even unwind from within a view controller embedded in a navigation controller when the view controller you're unwinding to has the navigation controller presented on top of it. (i.e. it will do a dismissViewController instead of a pop to unwind)
2) The Custom unwind method
Say you don't want to or cant trigger this action from a storyboard. There is still an option and its detailed over at this question here:
Whats the programmatic opposite of showViewController:sender:
The gist is you can write your own generic dismiss method by implementing categories on the UIKit container View controllers (or on your own container)

iOS Storyboards: How to segue to the second view with a back button to the first view, but without displaying the first view

Here is my storyboard configuration:
Navigation Controller -> View Controller A -> Push-> View Controller B
^
|
Modal
^
|
View Controller C
What I want to achieve: When a button is pressed in View C, directly View B will be opened modally (No part of View A is to be displayed). Also, View B will have a navigation back button to View A.
To achieve this,
I set up the illustrated storyboard.
I created a segue between View C and the Navigation Controller of View A/B.
In the 'prepareForSegue' method of View Controller C, I get an instance of View Controller A as the first element in the navigation. In this instance, I set a variable like 'directlyProceedToViewB=YES'.
In the viewDidLoad method of View Controller A, I check the variable 'directlyProceedToViewB' and if it is YES, I call 'performSegueWithIdentifier' to segue to View B
The result is so that, first View A is opened modally and after displaying it a very short time, View B is opened with a push animation (View B can navigate back to View A, which is good). But I do not want View A to be displayed any time at all. How can I achieve this?
EDIT:
To better visualize, I'm adding a screenshot with more example cases to support:
Here are some cases I want to support:
We can start with ViewC, click on 'Modally Display B' which opens ViewB, then click 'Back to A' to navigate back to ViewA, then click on 'Dismiss Modal' on ViewA to go back to ViewC
We can start with ViewD, clcik on 'Modally Display A' which opens ViewA, then click on 'PushB' to open ViewB, then go back and forth between A and B and modally dismiss to ViewD.
First of all, some corrections: those are not views but view controllers. And "view A" is not pushed into the UINavigationController but it's the root.
After that, I suggest making the segue in "view C" an unwind segue and implement the IBAction in "view A" by pushing "view B" with [[self navigationController] pushViewController:bViewController animated:NO].
EDIT (adding some details):
I assume that in ViewControllerA's viewWillAppear you present ViewControllerC in a not animated manner.
Implement an unwinding action like (IBAction)unwindAndThenGoToB:(UIStoryboardSegue *)segue in ViewControllerA.
In the storyboard connect the button in ViewControllerC to the Exit icon and select the previously defined method.
Then implement the method with the push call I wrote earlier.
ps: for documentation there is plenty on Apple's website.
Implement this using delegates.Decalre protocol in which class you want and define those methods and call the methods in the view controller you want.There is no many ways of calling some view and showing back button to go different view.modal view is just a concept.and you can use delegate methods to call whatever class you want.
Here I got a way to do so:-
You need to set no animation for segue from viewC to viewA as shown in below image. Then set a segue identifier for segue from viewA to viewB namely, "viewB" and in your viewA .m file add following code,
- (void)viewDidLoad {
[super viewDidLoad];
// Place your conditional check here.
[self performSegueWithIdentifier:#"viewB" sender:self]; //Will directly lead to viewB and viewA won't be shown as no animation is there from viewC to viewA.
}
And your rest flow be like-wise.
I found the solution myself.
First, I discovered that, my original proposal of
In the viewDidLoad method of View Controller A, I check the variable
'directlyProceedToViewB' and if it is YES, I call
'performSegueWithIdentifier' to segue to View B
works as I desired on iOS 7 but does not work on iOS 8.
So the solution is, in the viewDidLoad method of View Controller A, if 'directlyProceedToViewB' is YES, rather than calling performSegueWithIdentifier, use the following code:
ViewControllerB *destVC = [self.storyboard instantiateViewControllerWithIdentifier:#"ViewControllerBStoryboardID"];
[self.navigationController pushViewController:destVC animated:NO];

Creating a central log out view controller in storyboard

I am working with Parse, and one thing I have implemented in my app is their built in PFLogInViewController. This controller will be presented at two times in the application - when the app first starts and the user is not logged in, and when the user taps the "Log out" button of my application (logging out takes them back to the PFLogInViewController, as you are required to sign in to use the app). I would like to set this up using Storyboard, as that is how the rest of my app is laid out. How could I set up a central view controller (a PFLogInViewController) that is accessed at these two times? I have already Subclassed PFLogInViewController and set it up, I just need advice on how to place it in Storyboard and how to connect it to my views. To make this question help as many people as possible, the general theme of my question is how does one establish a central Login/ViewController that can be accessed at different points in the application using Storyboard. Attached is the basic idea of what I'm trying to accomplish. I haven't been able to successfully segue to the initial TabBarController, and I'm not sure how I should make the LoginController the initial ViewController if I can't segue. I am programming in Swift, if it matters.
There are a few ways to do this depending upon your application. One way is drop a UIViewController onto the storyboard, but don't wire it up to anything (no segue). Create a storyboard id for it such as "MyLoginVC". Do the necessary subclassing of UIViewController and attach the class to your VC. Then, when you want to display the VC simply do the following or wire this up to your logout button
id destinationVC = [self.storyboard instantiateViewControllerWithIdentifier:#"MyLoginVC"];
[self.navigationController pushViewController:destinationVC animated:YES];
In addition, if you want to show the login VC as the very first VC when you launch your app, then perhaps in your AppDelegate
// Load Root view controller
UIStoryboard *storyboard = [UIStoryboard storyboardWithName:#"Main" bundle:nil];
self.rootVC = [storyboard instantiateInitialViewController];
self.window.rootViewController = _rootVC;
[self.window makeKeyAndVisible];
// Load Login view controller
id initialVC = [storyboard instantiateViewControllerWithIdentifier:#"MyLoginVC"];
[initialVC setModalPresentationStyle:UIModalPresentationFullScreen];
[_rootVC presentModalViewController:initialVC animated:NO];
When you finish with your login VC (i.e. successful login) then within login VC
[self dismissViewControllerAnimated:NO completion:nil];
and alternatively instantiate your first VC with something similar to the following from within login VC. Note, since you loaded the root VC above first, it is already there with the login VC sitting over it. When you dismiss login VC, the underlying root VC should be ready to rock and roll. Otherwise you can do the following:
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
RootTabBarController *tbController = (RootTabBarController *)[self.storyboard instantiateViewControllerWithIdentifier:#"rootTabBarController"];
[self presentViewController:tbController animated:YES completion:NULL];
}
I think what you want is an unwind segue. Here are the instructions I follow for an unwind segue: https://github.com/bradley/iOSUnwindSegueProgramatically
If the link dies, here is what it said:
In your storyboard create two view controllers.
Subclass UIViewController twice, once for each of the view controllers in your storyboard.
Connect these view controllers to the view controllers in your storyboard.
Make a segue between the first view controller and the second by control+dragging from the first to the second.
Click on the segue you created and give it an identifier in the attributes inspector.
Make a button on the first view controller and link it to an IBAction in its UIViewController subclass.
When this button is pressed, the second storyboard should appear. To make this happen (we are doing it programatically) put the following into the implementation of the action you just created:
[self performSegueWithIdentifier:#"nameOfTheSegueBetweenOneAndTwo" sender:self];
Create a second method in the implemention of the first view controller with the following:
- (IBAction)returnToStepOne:(UIStoryboardSegue *)segue {
NSLog(#"And now we are back.");
}
This method will work to unwind any view controller back to this view controller. Notice that we implement the method in the view controller we wish to return to.
Go back to the storyboard. Focus in on the second view controller. If it is active, you should see a dark bar beneath it with 3 symbols on it. One of these is orange and when hovered over will show the name of the UIViewController subclass that this view controller represents. Control drag from this symbol woth the green symbol that means 'Exit'. You should see all available segue unwinds, which XCode automatically enumerates when you create segue unwind implementations inside UIViewController subclasses that you have shown on your stroryboard. Hence, you should see the segue 'returnToStepOne' as an option. Select it.
In your storyboard's document outline, find the section for the second view controller. You should see an item listed below it with a grey symbol that says something like "Unwind segue from ... to Exit." Click on this item.
Important and easily missed step follows!
On the right side of your storyboard, in the attributes inspector, you should see two fields. One for 'Identifier' and one for 'Action'. In most cases, the 'Action' field will have the text 'returnToStepOne:', which is what we want, but the 'Identifier' field will be blank. Fill this field with the text: 'returnToStepOne' (note that we leave out the colon).
Create a button on the second view controller and link it to an IBAction in its UIViewController subclass.
In the implementation for the method you just created, put the following code:
[self performSegueWithIdentifier:#"returnToStepOne" sender:self];
Run the application. You should now be able to unwind from the second view controller to the first.

What is the proper way to dismiss a modal when using storyboards?

Using storyboards, what is the proper way to dismiss a modal?
using IBAction and writing code to dismiss after a button click?
using segue and notify the parent view controller after a button click?
See Here Dismissing a Presented View Controller about halfway down
When it comes time to dismiss a presented view controller, the preferred approach is to let the presenting view controller dismiss it.
So you should use an IBAction and writing code to dismiss after a button click
According Alex Cio answer for Swift 3 and XCode 8.3:
Create class:
import UIKit
class DismissSegue: UIStoryboardSegue {
override func perform() {
self.source.presentingViewController?.dismiss(animated: true, completion: nil)
}
}
But in storyboard you should choose:
Action Segue -> Custom -> dismiss
Only after this option appear on Action Segue menu
I've found that usually when I'm attempting to do this in storyboard I'd rather not create extra classes. It still makes sense to perform the dismiss from the presenting view controller, so that requires a class to back it.
If you create an IBAction in the presenting view controller and name it appropriately e.g.
- (IBAction)dismissAnyModel:(id)sender
{
[self dismissViewControllerAnimated:YES completion:nil];
}
Then from storyboard wherever you want to trigger the dismiss from you create an action to the first responder as shown below. You can extend this to work with multiple presenting view controllers by creating unique names for the IBActions.
More information on first responder and the responder chain
See my answer here. It gives you two ways to dismiss the modal view controller with storyboard. I like method two described because one you add the class in your project your return from modal views can be done with no code using storyboard alone. That said, if you have implemented a delegate and delegate protocol, it is also a good place to put the dismissModalViewController statement.
To do this inside the UIStoryboard you need first to create an Object of the type UIStoryboardSegue in your project
Then insert following method inside the class. Here is my class
#implementation DismissController
- (void)perform{
UIViewController *sourceVC = self.sourceViewController;
[sourceVC.presentingViewController dismissViewControllerAnimated:YES
completion:nil];
}
Now you can use it inside your UIStoryboard. Select the button that should make the UIViewController Disappear and drag it to the UIViewController you want to go to. In my case it shows **dismiss Controller* because of the name of my Class.
Select it and you are done!
There is also a very good explanation on this website.
As the Apple online documentation indicates, the presenting view controller is responsible for dismissing the modal (presented) view.
There's a post and example available
here

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