我想使用Swift代码在我的应用程序中正确定位项目,无论屏幕大小如何。例如,如果我想要一个按钮是屏幕宽度的75%,我可以这样做:(screenWidth * .75)
是按钮的宽度。我发现这可以在Objective-C中通过执行以下操作来确定
CGFloat screenWidth = screenSize.width;
CGFloat screenHeight = screenSize.height;
不幸的是,我不确定如何将其转换为Swift。有谁有主意吗?
谢谢!
发布于 2014-06-09 05:05:41
Swift 5.0中的
let screenSize: CGRect = UIScreen.main.bounds
Swift 4.0
// Screen width.
public var screenWidth: CGFloat {
return UIScreen.main.bounds.width
}
// Screen height.
public var screenHeight: CGFloat {
return UIScreen.main.bounds.height
}
Swift 3.0中的
let screenSize = UIScreen.main.bounds
let screenWidth = screenSize.width
let screenHeight = screenSize.height
旧版swift中的:
这样做:
let screenSize: CGRect = UIScreen.mainScreen().bounds
然后,您可以像这样访问宽度和高度:
let screenWidth = screenSize.width
let screenHeight = screenSize.height
如果你想要75%的屏幕宽度,你可以去:
let screenWidth = screenSize.width * 0.75
https://stackoverflow.com/questions/24110762
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