Swift, iOS 9/10 (UIPopoverController弃用后)

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

    if UIDevice.currentDevice().userInterfaceIdiom == .Pad {

       if activityViewController.respondsToSelector(Selector("popoverPresentationController")) {
          activityViewController.popoverPresentationController?.sourceView = self.view
        }
    }

    self.presentViewController(activityViewController, animated: true, completion: nil)

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;
 }

我找到了这个解 首先，你呈现弹窗的视图控制器应该实现<UIPopoverPresentationControllerDelegate>协议。

接下来，你需要设置popoverPresentationController的委托。

添加以下功能:

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
// Assuming you've hooked this all up in a Storyboard with a popover presentation style
    if ([segue.identifier isEqualToString:@"showPopover"]) {
        UINavigationController *destNav = segue.destinationViewController;
        PopoverContentsViewController *vc = destNav.viewControllers.firstObject;

        // This is the important part
        UIPopoverPresentationController *popPC = destNav.popoverPresentationController;
        popPC.delegate = self;
    }
}

- (UIModalPresentationStyle)adaptivePresentationStyleForPresentationController: (UIPresentationController *)controller {
    return UIModalPresentationNone;
}

解决Objective-C和使用UIPopoverPresentationController

    UIActivityViewController *controller = /*Init your Controller*/;
    //if iPhone
    if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone) {
        [self presentViewController:controller animated:YES completion:nil];
    }
    //if iPad
    else {
        UIPopoverPresentationController* popOver = controller.popoverPresentationController
        if(popOver){
            popOver.sourceView = controller.view;
            popOver.sourceRect = CGRectMake(self.view.frame.size.width/2, self.view.frame.size.height/4, 0, 0);
            [self presentViewController:controller animated:YES completion:nil];
        }
    }

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)
        }

    }
}

我尝试了下一个代码，它工作:

首先在视图控制器中放入一个栏按钮项 然后创建一个IBOutlet:

@property(weak,nonatomic)IBOutlet UIBarButtonItem *barButtonItem;

你的uiactivityviewcontroller . popoverpresentationcontroller . barbuttonitem = self.barButtonItem;