Strange Custom Background Color on UIPickerView Swift

Question

I'm getting strange colors when assigning a custom UIColor to the background of UIPickerViews. I have created a color for textViews and pickerViews as follows:

let myTextViewBackgroundColor = UIColor(red: 192.0/255.0, green: 255.0/255.0, blue: 255.0/255.0, alpha: 0.35)

And I assign it as:

myTextView.backgroundColor = myTextViewBackgroundColor
keywordPicker.backgroundColor = myTextViewBackgroundColor

Here's what it looks like: As you can see, the color of the textViews and the color of the picker are different. The difference seems more pronounced on a device than in this snapshot, but the custom color looks muddy, almost like it is mixed with a default gray background.

If I comment out the pickerView color item above, I get a gray color background: If I choose a standard UIColor, it appears to work correctly. UIColor.yellowColor in this instance. If I choose white, that looks good, too. I create the pickerView in code and assign it to a textField which has no background color assigned.

var keywordPicker : UIPickerView?

keywordPicker = UIPickerView()
keywordPicker?.delegate = self
keywordPicker?.dataSource = self

keywordsForItemTextField.inputView = keywordPicker
keywordPicker.backgroundColor = myTextViewBackgroundColor

And for what it's worth, I have a date picker that I created as a modal screen with it's own view controller. In that case, the picker is created on the storyboard. I still assign my background color with code and that picker shows the correct color.

Answer 1

tkanzakic's answer finally allowed me to change the background color of UIDatePicker.

Here's the Swift 4 version:

class CustomDatePicker: UIDatePicker {

    var customBackgroundColor = UIColor.black

    override func willMove(toWindow newWindow: UIWindow?) {
        super.willMove(toWindow: newWindow)

        if newWindow != nil {
            inputView?.backgroundColor = customBackgroundColor
        }
    }
}

Change the customBackgroundColor property to your liking.