无论屏幕大小如何,我都想使用Swift代码在我的应用程序中正确定位项目。例如,如果我希望按钮(screenWidth *.75)的宽度为屏幕宽度的75%,则可以做为按钮的宽度。我发现可以通过在Objective-C中确定
(screenWidth *.75)
CGFloat screenWidth = screenSize.width; CGFloat screenHeight = screenSize.height;
不幸的是,我不确定如何将其转换为Swift。有人有主意吗?
谢谢!
在Swift 3.0中
let screenSize = UIScreen.main.bounds let screenWidth = screenSize.width let screenHeight = screenSize.height
快点做:做这样的事情:
let screenSize: CGRect = UIScreen.mainScreen().bounds
那么您可以像这样访问宽度和高度:
let screenWidth = screenSize.width let screenHeight = screenSize.height
如果您想要屏幕宽度的75%,则可以执行以下操作:
let screenWidth = screenSize.width * 0.75
迅捷4.0
// Screen width. public var screenWidth: CGFloat { return UIScreen.main.bounds.width } // Screen height. public var screenHeight: CGFloat { return UIScreen.main.bounds.height }
在Swift 5.0中
let screenSize: CGRect = UIScreen.main.bounds