通过 UIActivityViewController 分享到 Twitter/Facebook 等导致崩溃

Question

在 iOS8 上，我使用 UIActivityViewController 将 UIImage 共享到 Facebook/Twitter 等。它似乎工作正常，但今天在我的 iPad 上运行代码时突然开始崩溃。但是，它仍然可以在模拟器中按预期工作。

我的代码：

UIActivityViewController *controller =
[[UIActivityViewController alloc]
 initWithActivityItems:@[text, url, myImage]
 applicationActivities:nil];

[self presentViewController:controller animated:YES completion:nil];

在崩溃时，Xcode 吐出：

发现的扩展：{( {id = com.apple.share.Facebook.post}， {id = com.apple.share.Twitter.post}， {id = com.apple.share.TencentWeibo.post}, {id = com.apple.share.SinaWeibo.post} )} 用于属性：{ NSExtensionActivationRule = { 扩展项 = ( { 附件 = ( { 注册类型标识符 = ( “公众形象” ); }, { 注册类型标识符 = ( “public.plain-text” ); }, { 注册类型标识符 = ( “公共.url” ); } ); } ); }; NSExtensionPointName = ( "com.apple.share-services", "com.apple.ui-services", “com.apple.services” ); 2014-08-07 21:38:59.208 collageTest[279:11021] LaunchServices：invalidationHandler 称为 2014-08-07 21:38:59.212 collageTest[279:11016] 发现的扩展：{( {id = com.apple.share.Flickr.post}， {id = com.apple.mobileslideshow.StreamShareService}， {id = com.apple.share.Twitter.post}， {id = com.apple.share.Facebook.post}， {id = com.apple.share.Vimeo.post}， {id = com.apple.share.SinaWeibo.post}, {id = com.apple.share.TencentWeibo.post} )} 属性：{ NSExtensionPointName = "com.apple.share-services"; 2014-08-07 21:38:59.216 collageTest[279:11021] LaunchServices： 调用了 invalidationHandler

Answer 1

查看the docs，我需要为弹出框控制器定义一个源视图

UIActivityViewController *controller =
[[UIActivityViewController alloc]
 initWithActivityItems:@[text,url,myImage]
 applicationActivities:nil];

[self presentViewController:controller animated:YES completion:nil];

UIPopoverPresentationController *presentationController =
[controller popoverPresentationController];

presentationController.sourceView = self.view;

Answer 2

popoverPresentationController 是 iOS 8 的新成员，将在 iOS 7 上崩溃。它在 iPhone 上也为零，因为它仅在 iPad 上的 UIPopover 中。以下是 Christian 在 Swift 中的回答，并考虑了这些事实：

Swift 2.0 (Xcode 7)

let controller = UIActivityViewController(activityItems: [text, url, myImage], applicationActivities: nil)

presentViewController(controller, animated: true, completion: nil)

if #available(iOS 8.0, *) {
    let presentationController = controller.popoverPresentationController
    presentationController?.sourceView = view
}

Swift 1.2 (Xcode 6)

let controller = UIActivityViewController(activityItems: [text, url, myImage], applicationActivities: nil)

presentViewController(controller, animated: true, completion: nil)

if controller.respondsToSelector("popoverPresentationController") {
    // iOS 8+
    let presentationController = controller.popoverPresentationController
    presentationController?.sourceView = view
}

Answer 3

正如@mmccomb 告诉here，在 iPad 上，活动视图控制器将使用新的 UIPopoverPresentationController 显示为弹出框。您至少需要指定源视图：

activityViewController.popoverPresentationController.sourceView = YOURDESIREDVIEW;

如果要显示锚定到该视图任意点的弹出框，请使用 popoverPresentationController 的 sourceRect 属性指定它。

Answer 4

这是我用 swift 解决的方法：

    let someText:String = "shareText"
    let google:NSURL = NSURL(string:"http://google.com")!

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

    if respondsToSelector("popoverPresentationController") {
        self.senderView.presentViewController(activityViewController, animated: true, completion: nil)
        activityViewController.popoverPresentationController?.sourceView = sender
    }else{
        senderView.presentViewController(activityViewController, animated: true, completion: nil)
    }

Answer 5

Swift 5 - 从工作项目复制

问题

如果您通过以下方式呈现UIActivityViewController 进行分享，您的应用将正常运行，但会在 iPad 设备上崩溃。

func shareSiteURL()
{
    let title = "My app"
    let url = URL(string: "https://myWebSite.com")

    let activityController = UIActivityViewController(activityItems: [url!, title], applicationActivities: nil)
    self.present(activityController, animated: true, completion: nil)
}

说明

当您在 iPad 上呈现UIActivityViewController 时，iPad 将活动控制器呈现为弹出框控制器，因此它需要知道和使用源视图。在前面的代码中，如果设备是 iPad，我们没有传递源视图，这就是应用在 iPad 上运行时崩溃的原因。

解决方案

为了解决这个问题，我们将检查控制器是否显示为弹出框（应用程序在 iPad 上运行），如果是，我们将传递源视图，否则应用程序将崩溃。

func shareSiteURL()
{
    let title = "My app"
    let url = URL(string: "https://myWebSite.com")

    // check if the controller is presented as a popover (the app is running on iPad)
    // and if so, we will pass the source view otherwise the app will crash.

    if let popoverController = activityController.popoverPresentationController
    {
        popoverController.sourceView = viewController.view
    }

    let activityController = UIActivityViewController(activityItems: [url!, title], applicationActivities: nil)
    self.present(activityController, animated: true, completion: nil)
}