How to remove all subviews?

Question

When my app gets back to its root view controller, in the viewDidAppear: method I need to remove all subviews.

How can I do this?

Warm

Answer 1

Edit: With thanks to cocoafan: This situation is muddled up by the fact that NSView and UIView handle things differently. For NSView (desktop Mac development only), you can simply use the following:

[someNSView setSubviews:[NSArray array]];

For UIView (iOS development only), you can safely use makeObjectsPerformSelector: because the subviews property will return a copy of the array of subviews:

[[someUIView subviews]
 makeObjectsPerformSelector:@selector(removeFromSuperview)];

Thank you to Tommy for pointing out that makeObjectsPerformSelector: appears to modify the subviews array while it is being enumerated (which it does for NSView, but not for UIView).

Please see this SO question for more details.

Note: Using either of these two methods will remove every view that your main view contains and release them, if they are not retained elsewhere. From Apple's documentation on removeFromSuperview:

If the receiver’s superview is not nil, this method releases the receiver. If you plan to reuse the view, be sure to retain it before calling this method and be sure to release it as appropriate when you are done with it or after adding it to another view hierarchy.

Bugger

Are you sure this is safe? It mutates the list while iterating it, and I'm unable to find a definitive statement in Apple's documentation. – Schmitt

@Tommy: That is a good point. Some Googling turned up the answer: UIView returns a copy of the subviews mutable array, so this code just works. Completely different story on the desktop, where the same code will throw an exception. See #4665679 – Bugger

@Tommy: I have updated my answer accordingly. Thank you for catching that! – Bugger

UIView does not respond to setSubviews:, does it? – Cita

@cocoafan: It looks like you are right. You can only set the subviews from within subclasses of UIView. It is NSView that allows setting them externally. – Bugger

Couldn't find a way to do this in Swift – Footrace

the Xamarin way : someUIView.Subviews.All(p => p.RemoveFromSuperview); – Latticework

Good answer, Slightly different way: create a UIView category and implement some -(void)removeSubviews method. Same mechanism, but prettier code. – Candlemaker

@BenoitJadinon - won't compile - you appear to mean abusing All to perform ForEach, so someUIView.Subviews.All( v => { v.RemoveFromSuperview(); return true; } );. IMHO cleaner to say what you mean: someUIView.Subviews.ToList().ForEach( v => v.RemoveFromSuperview() );. – Berni

Answer 2

Get all the subviews from your root controller and send each a removeFromSuperview:

NSArray *viewsToRemove = [self.view subviews];
for (UIView *v in viewsToRemove) {
    [v removeFromSuperview];
}

Dextrose

+1 and thank you. I should have also used self.view as you have. – Bugger

why not!? for (UIView *v in [self.view subviews]) its easier – Jannery

@Jannery It's much clearer and more verbose the way he did it. Verbose and readability > saving keystrokes – Cytolysin

@Cytolysin You should have said "readability is more important than saving keystrokes" instead of "readability > saving keystrokes" and then your comment would be more readable. :-) – Brodsky

Let's not forget about the fact that if [self.view subviews] performs any calculations under the hood, putting it directly in the for loop could cause those calculations to be performed over and over again. Declaring it before the loop ensures they are only performed once. – Umeko

Answer 3

In Swift you can use a functional approach like this:

view.subviews.forEach { $0.removeFromSuperview() }

As a comparison, the imperative approach would look like this:

for subview in view.subviews {
    subview.removeFromSuperview()
}

These code snippets only work in iOS / tvOS though, things are a little different on macOS.

Rickrickard

(subviews as [UIView]).map { $0.removeFromSuperview() } – Splenitis

it's not functional since a function returns a value and this just discards the result of the .map. this is a pure side effect and is better handled like this: view.subviews.forEach() { $0.removeFromSuperview() } – Marvin

You are right, Martin, I agree with you. I just didn't know there was a forEach() method on Arrays. When was it added or did I just oversee it? I've updated my answer! – Rickrickard

I am so lazy that even if I knew how to clear subviews, I still came here to copy/paste your snippet and put a +1 – Virgievirgil

Answer 4

If you want to remove all the subviews on your UIView (here yourView), then write this code at your button click:

[[yourView subviews] makeObjectsPerformSelector: @selector(removeFromSuperview)];

Alvy

Welcome to Stack Overflow! Would you consider adding some narrative to explain why this code works, and what makes it an answer to the question? This would be very helpful to the person asking the question, and anyone else who comes along. Additionally, the already-accepted answer includes code that is essentially the same as this. – Alginate

How could this help more so than the accepted answer: It's identical. Why write this? – Runstadler

Answer 5

This does only apply to OSX since in iOS a copy of the array is kept

When removing all the subviews, it is a good idea to start deleting at the end of the array and keep deleting until you reach the beginning. This can be accomplished with this two lines of code:

for (int i=mySuperView.subviews.count-1; i>=0; i--)
        [[mySuperView.subviews objectAtIndex:i] removeFromSuperview];

SWIFT 1.2

for var i=mySuperView.subviews.count-1; i>=0; i-- {
    mySuperView.subviews[i].removeFromSuperview();
}

or (less efficient, but more readable)

for subview in mySuperView.subviews.reverse() {
    subview.removeFromSuperview()
}

NOTE

You should NOT remove the subviews in normal order, since it may cause a crash if a UIView instance is deleted before the removeFromSuperview message has been sent to all objects of the array. (Obviously, deleting the last element would not cause a crash)

Therefore, the code

[[someUIView subviews] makeObjectsPerformSelector:@selector(removeFromSuperview)];

should NOT be used.

Quote from Apple documentation about makeObjectsPerformSelector:

Sends to each object in the array the message identified by a given selector, starting with the first object and continuing through the array to the last object.

(which would be the wrong direction for this purpose)

Wynny

Can you please make an example of what you are referring to ? Don't know what are you referring to as "element" And how would this elements be removed before calling removeFromSuperView ? – Brainstorming

But how can an instance of UIView be deleted while calling this method ? Do you mean removed from the subview array ? – Brainstorming

When removeFromSuperview finishes, the UIView will be removed from the array, and if there are no other living instances with a strong relation to the UIView, the UIView will also be deleted. This may cause an out of bound exception. – Wynny

Gotcha! Thank you. I think you are getting a copy of the subviews array on IOS. In any case, it would be a good idea to make a copy yourself if you want to remove subviews. – Brainstorming

@simpleBob - did you read the comments written in 2011 on the accepted answer? According to those comments, on iOS, [yourView subviews] returns a COPY of the array, therefore is safe. (NOTE that on OSX, what you say is correct.) – Berni

Cannot use mutating member on immutable value: 'subviews' is a get-only property – Methodize

Answer 6

Try this way swift 2.0

view.subviews.forEach { $0.removeFromSuperview() }

Torgerson

Don't you see the date answer date i am earlier? Why not paste my answer link into that answer? – Torgerson

Right... the answer was posted before yours however the forEach based solution was added after yours, I missed that. Apologies. – Holleran

Answer 7

view.subviews.forEach { $0.removeFromSuperview() }

Estas

While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value. – Bove

Very nice answer. – Expander

Answer 8

Using Swift UIView extension:

extension UIView {
    func removeAllSubviews() {
        for subview in subviews {
            subview.removeFromSuperview()
        }
    }
}

Structural

Answer 9

In objective-C, go ahead and create a category method off of the UIView class.

- (void)removeAllSubviews
{
    for (UIView *subview in self.subviews)
        [subview removeFromSuperview];
}

Assorted

Answer 10

Use the Following code to remove all subviews.

for (UIView *view in [self.view subviews]) 
{
 [view removeFromSuperview];
}

Adverse

Answer 11

In order to remove all subviews Syntax :

- (void)makeObjectsPerformSelector:(SEL)aSelector;

Usage :

[self.View.subviews makeObjectsPerformSelector:@selector(removeFromSuperview)];

This method is present in NSArray.h file and uses NSArray(NSExtendedArray) interface

Scoutmaster

Answer 12

If you're using Swift, it's as simple as:

subviews.map { $0.removeFromSuperview }

It's similar in philosophy to the makeObjectsPerformSelector approach, however with a little more type safety.

Mansized

This is semantically incorrect, map should not result in side effects. Plus, the same result can be achieved via forEach. – Holleran

Answer 13

For ios6 using autolayout I had to add a little bit of code to remove the constraints too.

NSMutableArray * constraints_to_remove = [ @[] mutableCopy] ;
for( NSLayoutConstraint * constraint in tagview.constraints) {
    if( [tagview.subviews containsObject:constraint.firstItem] ||
       [tagview.subviews containsObject:constraint.secondItem] ) {
        [constraints_to_remove addObject:constraint];
    }
}
[tagview removeConstraints:constraints_to_remove];

[ [tagview subviews] makeObjectsPerformSelector:@selector(removeFromSuperview)];

I'm sure theres a neater way to do this, but it worked for me. In my case I could not use a direct [tagview removeConstraints:tagview.constraints] as there were constraints set in XCode that were getting cleared.

Hovey

Answer 14

In monotouch / xamarin.ios this worked for me:

SomeParentUiView.Subviews.All(x => x.RemoveFromSuperview);

Muns

Answer 15

In order to remove all subviews from superviews:

NSArray *oSubView = [self subviews];
for(int iCount = 0; iCount < [oSubView count]; iCount++)
{
    id object = [oSubView objectAtIndex:iCount];
    [object removeFromSuperview];
    iCount--;
}

Kravits

Couple of major mistakes here @Pravin. First, you'd need 'object' to be defined as a UIView* otherwise you'd get a compiler error with [object removeFromSuperview]. Second, your for loop is already decrementing iCount so you are skipping an extra one with your iCount-- line. And finally, there are two working and correct approaches above and yours is neither more elegant nor faster. – Keenan

each iteration you do iCount++ and iCount--, leaving the index the same, so it will be an infinite loop if [oSubView count]>0. This is definitely buggy and NOT USABLE code. – Wynny