通过UIActivityViewController分享到Twitter / Facebook等导致崩溃

Question

在iOS8上，我正在使用UIActivityViewController与Facebook / Twitter等共享UIImage。它似乎工作正常，但今天它在我的iPad上运行代码时突然崩溃了。但是，它仍然可以在模拟器中正常工作。

我的代码：

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

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

崩溃后，Xcode吐出：

  

发现的扩展名：{（        {id = com.apple.share.Facebook.post}，        {id = com.apple.share.Twitter.post}，        {id = com.apple.share.TencentWeibo.post}，        {id = com.apple.share.SinaWeibo.post}）} for attributes：{       NSExtensionActivationRule = {           extensionItems =（                           {                   附件=（                                           {                           registeredTypeIdentifiers =（                               “公众形象”                           ）;                       }，                                           {                           registeredTypeIdentifiers =（                               “public.plain文本”                           ）;                       }，                                           {                           registeredTypeIdentifiers =（                               “public.url”                           ）;                       }                   ）;               }           ）;       };       NSExtensionPointName =（           “com.apple.share服务”，           “com.apple.ui服务”，           “com.apple.services”       ）; 2014-08-07 21：38：59.208 collageTest [279：11021] LaunchServices：invalidationHandler名为2014-08-07 21：38：59.212   collageTest [279：11016]发现的扩展名：{（        {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}）} for attributes：{       NSExtensionPointName =“com.apple.share-services”; 2014-08-07 21：38：59.216 collageTest [279：11021] LaunchServices：   invalidationHandler名为

Answer 1

查看the docs，我需要为popover控制器定义源视图

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的新功能，并且会在iOS 7上崩溃。它在iPhone上也只是零，因为它只能在iPad上UIPopover。这是基督徒在斯威夫特的答案，考虑到这些事实：

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显示为弹出窗口。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的任何点的弹出窗口，请使用popoverPresentationController的sourceRect属性指定它。

Answer 4

以下是我用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-从工作项目中复制

问题

如果您以以下方式显示UIActivityViewController以便共享，则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上展示UIActivityViewController时，iPad会将活动控制器显示为弹出控件，因此它需要了解和使用源视图。在前面的代码中，如果设备是iPad，我们没有传递源视图，这就是为什么该应用在iPad上运行时崩溃的原因。

解决方案

要解决此问题，我们将检查控制器是否显示为弹出窗口（应用程序在iPad上运行），如果是，则将通过源视图，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)
}