Removing all subviews from a view

fladnag · October 2008

Is there a way to remove all of a view's subviews instead of removing them individually by name like in the code below?

[myView1 removeFromSuperview];

RickMaddy · October 2008

What about:


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

fladnag · October 2008

That worked perfectly. Thank you.

jshake11 · October 2009

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

The above gave me the Warning message of "UIView may not respond to -countByEnumerationWithState:objects:count: And my code crashed when I tried to run it.

I used this and it worked for me:



// Assuming *view is already set up as a UIView...

for (int i = 0; i < [[self.view subviews] count]; i++ ) {
    [[[self.view subviews] objectAtIndex:i] removeFromSuperview];
}

panzeriti · December 2009

Neither of these worked for me.

This would not compile for me:

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

And this did not remove all subviews for some reason:

for (int i = 0; i < [[self.view subviews] count]; i++ ) {
    [[[self.view subviews] objectAtIndex:i] removeFromSuperview];
}

But what did work is as follows:

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

This method compiled and removed all subviews.

BrianSlick · December 2009

The first one failed because subviews is not a property of self, which is presumably your view controller. If this was a custom UIView subclass, it would probably work.

The second failed because it isn't approaching the problem correctly. If you are making the array smaller each time, you have to work backwards. If you remove the object at index 0, then the object that was at index 1 is now at index 0. But your next pass removes the object at index 1, and so on.

I'm a little surprised that the 3rd one worked, because you can't change the size of the array when doing fast enumeration. I'm going to guess that it works here because the subviews property returns a copy of the array, so you aren't messing with the actual subview array.

Somebody posted a cool array trick here a few weeks ago. I haven't tested this code, but it should be something like this:

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

DenVog · March 2010

panzeriti;152135 said:
Neither of these worked for me.

But what did work is as follows:
for (UIView *view in [self.view subviews]) {
    [view removeFromSuperview];
}
This method compiled and removed all subviews.

Thanks for sharing this. I ran into the same issue. Your post save me a lot of time.

mndhkr · February 2011


while([view.subviews count] > 0)
   [[view.subviews lastObject] removeFromSuperview];

i think that a for cycle may cause some problem with the array varying in count while removing the subviews...

TUX2K · February 2011

mndhkr;303781 said:
while([view.subviews count] > 0)
   [[view.subviews lastObject] removeFromSuperview];
i think that a for cycle may cause some problem with the array varying in count while removing the subviews...

And thats what BrianSlick said.

mariano_donati · February 2011

Just to add another way:


for (int i = 0; i < [[self.view subviews] count]; i++ ) {
    [[[self.view subviews] objectAtIndex:0] removeFromSuperview];
}

DAW · April 2011

Verbose but easy to follow...

    // Clear up existing scale if we need to by removing any labels
    int i,j;
    int totalSubViews = [[self.view subviews] count];
    for (i=0,j=0; i < totalSubViews; i++ ) 
    {
        if ([[[self.view subviews] objectAtIndex:j] isKindOfClass:[UILabel class]])
        {
            UILabel *thisLabel = [[self.view subviews] objectAtIndex:j];
            NSLog (@\"Removing %@\",thisLabel.text);
            [thisLabel removeFromSuperview];
        }
        else
            j++;
    }

qdvictory · April 2011

BrianSlick;152192 said:
The first one failed because subviews is not a property of self, which is presumably your view controller. If this was a custom UIView subclass, it would probably work.

The second failed because it isn't approaching the problem correctly. If you are making the array smaller each time, you have to work backwards. If you remove the object at index 0, then the object that was at index 1 is now at index 0. But your next pass removes the object at index 1, and so on.

I'm a little surprised that the 3rd one worked, because you can't change the size of the array when doing fast enumeration. I'm going to guess that it works here because the subviews property returns a copy of the array, so you aren't messing with the actual subview array.

Somebody posted a cool array trick here a few weeks ago. I haven't tested this code, but it should be something like this:
[[[self view] subviews] makeObjectsPerformSelector: @selector(removeFromSuperview)];

this is a good way.

pygmybutcher · July 2011

BrianSlick;152192 said:
Somebody posted a cool array trick here a few weeks ago. I haven't tested this code, but it should be something like this:
[[[self view] subviews] makeObjectsPerformSelector: @selector(removeFromSuperview)];

This is the most elegant of the methods posted here. Thanks, Brian.

dany_dev · July 2011

really I prefer

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

:D

CBGraham · July 2011

pygmybutcher;350990 said:
This is the most elegant of the methods posted here. Thanks, Brian.

Elegant is a dirty word. makeObjectsPerformSelector says in the docs that you shouldn't give it something that modifies the array. removeFromSuperview does. Make a copy of the subviews array and work on it and stop trying to write one line solutions. (Same problem as a vanilla For loop affecting the array it is working over.)

BrianSlick · July 2011

CBGraham;360066 said:
Elegant is a dirty word. makeObjectsPerformSelector says in the docs that you shouldn't give it something that modifies the array. removeFromSuperview does. Make a copy of the subviews array and work on it and stop trying to write one line solutions. (Same problem as a vanilla For loop affecting the array it is working over.)

If you had actually read the documentation before posting, then you would know that this is unnecessary, since subviews is a copy property. Plus it is an NSArray, not an NSMutableArray, therefore it cannot be modified. So if this solution didn't work, it would crash very quickly, and this would have been noted here in this thread quite some time ago.

sneaky · August 2011

What about when I have views within views?
Will


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

remove all subviews of subviews or will I have to nest another for loop to get rid of them safely?

BrianSlick · August 2011

Perform experiments and find out.

sneaky · August 2011

BrianSlick;361843 said:
Perform experiments and find out.

That would make way too much sense, can't be doing that now! :D

So I set up a button to place a view as a subview to self.view. On this subview I put a button that calls the remove from superview for loop.

If the loop would not remove subviews of subviews I should see a leak in instruments but I was presenting and removing the view a good few times and no leak appeared so I think we can safely say that everything is indeed removed.

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