swift – 使用@ IBDesigable / @ IBInspectable在故事板中的按钮上不显示转角半径

import UIKit

@IBDesignable
class CustomButton: UIButton {

    @IBInspectable var cornerRadiusValue: CGFloat = 10.0 {
        didSet {
            setUpView()
        }
    }

    required init?(coder aDecoder: NSCoder) {
        super.init(coder: aDecoder)
        self.layer.cornerRadius = 10.0
    }

    override func awakeFromNib() {
        setUpView()
    }

    override func prepareForInterfaceBuilder() {
        super.prepareForInterfaceBuilder()
        setUpView()
    }

    func setUpView() {
        self.layer.cornerRadius = 10.0
    }

}

但是角落半径没有显示在故事板中的按钮上.

我理解@IBInspectable只允许您更改检查器面板中的值.我想这不是我想要的.

当我创建一个具有该类的按钮时,我希望角落半径只显示在故事板中.我认为这就是@IBDesignable的作用.

@IBDesignable
class CustomButton: UIButton {
    @IBInspectable var cornerRadiusValue: CGFloat = 10.0 {
        didSet {
            setUpView()
        }
    }
    override func awakeFromNib() {
        super.awakeFromNib()
        setUpView()
    }
    override func prepareForInterfaceBuilder() {
        super.prepareForInterfaceBuilder()
        setUpView()
    }
    func setUpView() {
        self.layer.cornerRadius = self.cornerRadiusValue
        self.clipsToBounds = true
    }
}

现在按钮既可检测又可设计：

它也适用于正在运行的应用程序.