Removing all CALayer's sublayers

Question

I have trouble with deleting all of layer\'s sublayers. I currently do this manually, but that brings unnecessary clutter. I found many topics about this in google, but no a

Answer 1

The following should work:

for (CALayer *layer in [[rootLayer.sublayers copy] autorelease]) {
    [layer removeFromSuperlayer];
}

Answer 2

Calling rootLayer.sublayers = nil; can cause a crash (e.g. if, under iOS 8, you call removeFromSuperview twice on the view owning rootLayer).

The right way should be:

[[rootLayer.sublayers copy] makeObjectsPerformSelector:@selector(removeFromSuperlayer)]

The call to copy is needed so that the array on which removeFromSuperlayer is iteratively called is not modified, otherwise an exception is raised.

Answer 3

[rootLayer.sublayers makeObjectsPerformSelector:@selector(removeFromSuperlayer)];

Answer 4

I had to do this in Xamarin/C#. I had a UITableViewCell with some CAShapeLayers for borders. All of the above options (including copying the Array and then removing layers caused a crash). The approach that worked for me:

When adding the CALayer, I gave it a name:

    var border = new CALayer();
    border.BackgroundColor = color.CGColor;
    border.Frame = new  CGRect(x, y, width, height);
    border.Name = "borderLayerName";
    Layer.AddSublayer(border);

In PrepareForReuse on the UITableViewCell, I created a copy of the SubLayers property and removed anything that matched the name I assigned earlier:

    CALayer[] copy = new CALayer[Layer.Sublayers.Length];
    Layer.Sublayers.CopyTo(copy, 0);
    copy.FirstOrDefault(l => l.Name == "borderLayerName")?.RemoveFromSuperLayer();

No crashes.

Hope this helps!

Answer 5

You could simply provide an identifier for the CAlayer you have added and remove it by searching for it. Like so

 let spinLayer = CAShapeLayer()
 spinLayer.path = UIBezierPath()
 spinLayer.name = "spinLayer"


  func removeAnimation() {
     for layer in progressView.layer.sublayers ?? [CALayer]()
         where layer.name == "spinLayer" {
           layer.removeFromSuperlayer()
       }
   }

Answer 6

What about doing:

rootLayer.sublayers = @[];