Swift creating UIButton programmatically with dynamic height

Question

I'm wondering how can I create a UIButton programmatically with a dynamic height depending of the device. Let's say that I create the button like this:

bigButton = BigButton(frame: CGRect(x: 0, y: 0, width: self.view.frame.width , height: 60))

When setting the height, I need to put an integer, but I would like this to be 60px if display is for iPhone 6, 50px if iPhone 5... and so on.

I know how to do that if I create the button with IB, using the constraints and I could use a percentage height proportion, but since I need to create the button programmatically I'm kind of stuck with that. Thanks in advance!

Why not just add a height constraint programmatically? – Aaron Brager May 28 '16 at 16:18 — Aaron Brager, May 28 '16 at 16:18

score 1 · Accepted Answer · edited May 23 '17 at 11:52

Here's an example of something you can do:

var dynamicHeight: CGFloat
switch UIDevice().type {
    case .iPhone4:
        dynamicHeight = 40
    case .iPhone5:
        dynamicHeight = 50
    case .iPhone5S:
        dynamicHeight = 50
    case .iPhone6:
        dynamicHeight = 60
    case .iPhone6plus:
        dynamicHeight = 70
    default:
        dynamicHeight = 40
}

bigButton = BigButton(frame: CGRect(x: 0, y: 0, width: self.view.frame.width , height: dynamicHeight))

The method of retrieving the device type was taken from iOS: How to determine iphone model in Swift?.

Swift creating UIButton programmatically with dynamic height

1 Answers1