要在延迟后执行函数或代码，请使用下一个方法

DispatchQueue.main.asyncAfter(deadline: .now() + 'secondsOfDelay') {
        your code here...
    }

在本例中，函数getShowMovies将在1秒后执行

DispatchQueue.main.asyncAfter(deadline: .now() + 1) {
        self.getShowMovies()
    }

1)添加这个方法作为UIViewController Extension的一部分。

extension UIViewController{
func runAfterDelay(delay: NSTimeInterval, block: dispatch_block_t) {
        let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
        dispatch_after(time, dispatch_get_main_queue(), block)
    }
}

在VC上调用这个方法:

    self.runAfterDelay(5.0, block: {
     //Add code to this block
        print("run After Delay Success")
    })

2)

performSelector("yourMethod Name", withObject: nil, afterDelay: 1)

3)

override func viewWillAppear(animated: Bool) {

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, 2), dispatch_get_main_queue(), { () -> () in
    //Code Here
})

/ /紧凑的形式

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, 2), dispatch_get_main_queue()) {
    //Code here
 }
}

Matt的语法非常好，如果你需要使块失效，你可能想使用这个:

typealias dispatch_cancelable_closure = (cancel : Bool) -> Void

func delay(time:NSTimeInterval, closure:()->Void) ->  dispatch_cancelable_closure? {

    func dispatch_later(clsr:()->Void) {
        dispatch_after(
            dispatch_time(
                DISPATCH_TIME_NOW,
                Int64(time * Double(NSEC_PER_SEC))
            ),
            dispatch_get_main_queue(), clsr)
    }

    var closure:dispatch_block_t? = closure
    var cancelableClosure:dispatch_cancelable_closure?

    let delayedClosure:dispatch_cancelable_closure = { cancel in
        if closure != nil {
            if (cancel == false) {
                dispatch_async(dispatch_get_main_queue(), closure!);
            }
        }
        closure = nil
        cancelableClosure = nil
    }

    cancelableClosure = delayedClosure

    dispatch_later {
        if let delayedClosure = cancelableClosure {
            delayedClosure(cancel: false)
        }
    }

    return cancelableClosure;
}

func cancel_delay(closure:dispatch_cancelable_closure?) {

    if closure != nil {
        closure!(cancel: true)
    }
}

使用方法如下

let retVal = delay(2.0) {
    println("Later")
}
delay(1.0) {
    cancel_delay(retVal)
}

学分

上面的链接似乎坏了。来自Github的原始Objc代码

Swift 4有一个很短的方法来做到这一点:

Timer.scheduledTimer(withTimeInterval: 2, repeats: false) { (timer) in
    // Your stuff here
    print("hello")
}

斯威夫特 3+

这在Swift 3+中是超级简单和优雅的:

DispatchQueue.main.asyncAfter(deadline: .now() + 4.5) {
    // ...
}

年长的回答:

为了扩展Cezary的答案，它将在1纳秒后执行，我必须执行以下操作以在4秒半后执行。

let delay = 4.5 * Double(NSEC_PER_SEC)
let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay))
dispatch_after(time, dispatch_get_main_queue(), block)

编辑:我发现我原来的代码有一点错误。如果不将NSEC_PER_SEC转换为Double类型，隐式类型将导致编译错误。

如果有人能提出一个更优的解决方案，我很乐意听听。

Apple为Objective-C提供了dispatch_after代码段:

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, (int64_t)(<#delayInSeconds#> * NSEC_PER_SEC)), dispatch_get_main_queue(), ^{
    <#code to be executed after a specified delay#>
});

以下是移植到Swift 3的相同代码片段:

DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + <#delayInSeconds#>) {
  <#code to be executed after a specified delay#>
}