我看到很多人在使用Swift代码时对iPhone/iPad等进行硬编码。

这是不需要的，你必须使用语言功能。下面的代码假设你将使用一个UIBarButtonItem，并将在iPhone和iPad上工作。

@IBAction func share(sender: AnyObject) {
    let vc = UIActivityViewController(activityItems: ["hello"], applicationActivities: nil)
    vc.popoverPresentationController?.barButtonItem = sender as? UIBarButtonItem
    self.presentViewController(vc, animated: true, completion: nil)
 }

注意这里没有If语句或其他疯狂的东西。可选的展开在iPhone上将是nil，所以行vc。popoverpresentationcontroller ?在iphone上没有任何功能。

Swift = ios7/ ios8

let activityViewController = UIActivityViewController(activityItems: sharingItems, applicationActivities: nil)

//if iPhone
if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiom.Phone) {
    // go on..
} else {
    //if iPad
    if activityViewController.respondsToSelector(Selector("popoverPresentationController")) {
        // on iOS8
        activityViewController.popoverPresentationController!.barButtonItem = self.shareButtonItem;
    }
}
self.presentViewController(activityViewController, animated: true, completion: nil)

在Swift修复这个iPad，最好的方法是这样做，我发现。

    let things = ["Things to share"]
    let avc = UIActivityViewController(activityItems:things, applicationActivities:nil)
    avc.setValue("Subject title", forKey: "subject")
    avc.completionWithItemsHandler = {
        (s: String!, ok: Bool, items: [AnyObject]!, err:NSError!) -> Void in
    }

    self.presentViewController(avc, animated:true, completion:nil)
    if let pop = avc.popoverPresentationController {
        let v = sender as! UIView // sender would be the button view tapped, but could be any view
        pop.sourceView = v
        pop.sourceRect = v.bounds
    }

对于Swift 2.0。我发现，如果你试图将弹出窗口锚定在iPad上的共享按钮上，这种方法是有效的。这假设您已经在工具栏中为共享按钮创建了一个出口。

func share(sender: AnyObject) {
    let firstActivityItem = "test"

    let activityViewController = UIActivityViewController(activityItems: [firstActivityItem], applicationActivities: nil)

    if UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiom.Phone {
        self.presentViewController(activityViewController, animated: true, completion: nil)
    }
    else {            
        if activityViewController.respondsToSelector("popoverPresentationController") {
            activityViewController.popoverPresentationController!.barButtonItem = sender as? UIBarButtonItem
            self.presentViewController(activityViewController, animated: true, completion: nil)
        }

    }
}

修复了Swift 2.0

    if UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiom.Phone {
        self.presentViewController(activityVC, animated: true, completion: nil)
    }
    else {
        let popup: UIPopoverController = UIPopoverController(contentViewController: activityVC)
        popup.presentPopoverFromRect(CGRectMake(self.view.frame.size.width / 2, self.view.frame.size.height / 4, 0, 0), inView: self.view, permittedArrowDirections: UIPopoverArrowDirection.Any, animated: true)
    }

在iPad上，活动视图控制器将显示为一个弹出窗口使用新的UIPopoverPresentationController，它要求你指定一个锚点为弹出窗口的表示使用以下三个属性之一:

栏按钮项 源视图 源矩形

为了指定锚点，你需要获得UIActivityController的UIPopoverPresentationController的引用，并设置其中一个属性如下所示:

if ( [activityViewController respondsToSelector:@selector(popoverPresentationController)] ) { 
// iOS8
 activityViewController.popoverPresentationController.sourceView =
parentView;
 }