Swift:确定iOS屏幕大小

问题描述

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，如果我希望按钮的宽度为屏幕宽度的75％，则可以执行(screenWidth * .75)这样的按钮宽度.我发现这可以通过在Objective-C中确定来完成

I would like to use Swift code to properly position items in my app for no matter what the screen size is. For example, if I want a button to be 75% of the screen wide, I could do something like (screenWidth * .75) to be the width of the button. I have found that this could be determined in Objective-C by doing

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?

Unfortunately, I am unsure of how to convert this to Swift. Does anyone have an idea?

谢谢！

推荐答案

在Swift 3.0中

let screenSize = UIScreen.main.bounds
let screenWidth = screenSize.width
let screenHeight = screenSize.height

以较早的速度:做这样的事情:

In older swift:Do something like this:

let screenSize: CGRect = UIScreen.mainScreen().bounds

然后您可以像这样访问宽度和高度:

then you can access the width and height like this:

let screenWidth = screenSize.width
let screenHeight = screenSize.height

如果您想要屏幕宽度的75％，则可以:

if you want 75% of your screen's width you can go:

let screenWidth = screenSize.width * 0.75

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
}

在Swift 5.0中

let screenSize: CGRect = UIScreen.main.bounds

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