make a portion of a UIView become transparent

Question

I have a UIView w/c I returned as view in my viewForHeaderInSection. Now, I want to make a small transparent square on the lower right of the view so the cells will be seen through. How can I achieve something like this? Do I need to use this CGContextSetBlendMode? can anyone shed light on this.

Answer 1

Here is a UIView subclass solution. In order for this to work you must leave the superviews backgroundColor property as nil witch is the default value. If this UIView-subclass is in the storyboard then make sure you set the background color to "Clear color" in the attributes inspector.

- (void)drawRect:(CGRect)rect {
    CGContextRef ctx = UIGraphicsGetCurrentContext();
    UIColor *blackBackgroundColor = [UIColor colorWithRed:0 green:0 blue:0 alpha:1];
    CGContextSetFillColorWithColor(ctx, blackBackgroundColor.CGColor);
    CGContextFillRect(ctx, rect);
    CGRect transparentPart = CGRectMake(10, 10, 120, 80);
    CGContextClearRect(ctx, transparentPart);
}

Answer 2

CALayer *layer = [[CALayer alloc] init];
[layer setFrame:yourBoundsInView];

[[yourView layer] setMask:layer];

or override the drawRect method and add the below lines of code at the end

UIBezierPath *seeThrough = [UIBezierPath bezierPathWithRect:seeThroughRect];
[[UIColor colorWithWhite:1 alpha:0] set];
[seeThrough fillWithBlendMode:kCGBlendModeSourceIn alpha:1];

Answer 3

You need to apply mask for your View.

Tutorial: https://medium.com/@peteliev/layer-masking-for-beginners-c18a0a10743

Answer 4

The view are inserted as subviews inside of SwitchViewController.view. So, they will always appear above the SwitchViewControllers view. The other view is removed, Only one view will be appeared at 1 time

eg Here

[yellowViewController.view removeFromSuperview];
[self.view insertSubview:blueViewController.view atIndex:0];

view appears behind its subview by default, and

insertSubview:atIndex:

can't place it behind the view itself, as a view is not contained in the list of its own subviews.