我目前正在用Xcode 6 (Beta 6)测试我的应用程序。UIActivityViewController在iPhone设备和模拟器上工作得很好,但在iPad模拟器和设备(iOS 8)上崩溃

Terminating app due to uncaught exception 'NSGenericException', 
reason: 'UIPopoverPresentationController 
(<_UIAlertControllerActionSheetRegularPresentationController: 0x7fc7a874bd90>) 
should have a non-nil sourceView or barButtonItem set before the presentation occurs.

我使用以下代码用于iPhone和iPad的iOS 7以及iOS 8

NSData *myData = [NSData dataWithContentsOfFile:_filename];
NSArray *activityItems = [NSArray arrayWithObjects:myData, nil];
UIActivityViewController *activityViewController = [[UIActivityViewController alloc] initWithActivityItems:nil applicationActivities:nil];
activityViewController.excludedActivityTypes = @[UIActivityTypeCopyToPasteboard];
[self presentViewController:activityViewController animated:YES completion:nil];

我得到一个类似的崩溃在我的另一个应用程序以及。你能引导我吗?ios8中的UIActivityViewController有什么变化吗?我查过了,但在这上面什么也没找到


当前回答

斯威夫特3:

class func openShareActions(image: UIImage, vc: UIViewController) {
    let activityVC = UIActivityViewController(activityItems: [image], applicationActivities: nil)
    if UIDevice.current.userInterfaceIdiom == .pad {
        if activityVC.responds(to: #selector(getter: UIViewController.popoverPresentationController)) {
            activityVC.popoverPresentationController?.sourceView = vc.view
        }
    }
    vc.present(activityVC, animated: true, completion: nil)
}

其他回答

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;
}

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)
        }

    }
}

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;
 }

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)

如果你是在iPad上使用swift进行开发,那么你就得小心了,它在调试时可以正常工作,但在发布时就会崩溃。为了让它与testFlight和AppStore一起工作,禁用swift的优化,使用-none进行发布。