将应用 link 分享给 ActivityViewController iOS swift?

Share app link to by ActivityViewController iOS swift?

我的应用程序尚未上线。我从 App Store Connect 获得了应用程序 ID。我想在社交媒体应用程序上分享应用程序 link。我使用了 UIActivityViewController:

let string1 = "itms-apps://itunes.apple.com/app/idXXXXXXX"
let url = NSURL(string: string1)

let shareItems = [UIApplication.sharedApplication().openURL(url!)]
    
let activityViewController = UIActivityViewController(activityItems: shareItems, applicationActivities: nil)
self.presentViewController(activityViewController, animated: true, completion: nil)

问题:它没有显示 WhatsApp 等社交媒体应用程序。

这是用来打开网站的,不是用来分享应用的:

[UIApplication.sharedApplication().openURL(url!)]

改为这样做:

if let name = URL(string: "https://itunes.apple.com/us/app/myapp/idxxxxxxxx?ls=1&mt=8"), !name.absoluteString.isEmpty {
  let objectsToShare = [name]
  let activityVC = UIActivityViewController(activityItems: objectsToShare, applicationActivities: nil)
  self.present(activityVC, animated: true, completion: nil)
} else {
  // show alert for not available
}

示例见this

Swift4 或更好的解决方案:

此解决方案也适用于 iPad(上述解决方案在 iPad 上崩溃):

if let urlStr = NSURL(string: "https://apps.apple.com/us/app/idxxxxxxxx?ls=1&mt=8") {
    let objectsToShare = [urlStr]
    let activityVC = UIActivityViewController(activityItems: objectsToShare, applicationActivities: nil)

    if UIDevice.current.userInterfaceIdiom == .pad {
        if let popup = activityVC.popoverPresentationController {
            popup.sourceView = self.view
            popup.sourceRect = CGRect(x: self.view.frame.size.width / 2, y: self.view.frame.size.height / 4, width: 0, height: 0)
        }
    }

    self.present(activityVC, animated: true, completion: nil)
}

iPhone Swift 5+

的解决方案
let url = URL(string: "https://apps.apple.com/us/app/id1535629801")!
let vc = UIActivityViewController(activityItems: [url], applicationActivities: nil)
present(vc, animated: true)

这里的解决方案都很好,但值得考虑实施 UIActivityItemSource 协议和 LinkPresentation 框架。

我的解决方案实现了以下目标:

  • 应用程序图标和标题显示在 UIActivityViewController
  • 直接 link 到 App Store 进行 AirDrop ActivityType
  • 消息和电子邮件的自定义文本,包括在 Google 上向应用添加 link 的机会 如果需要,请播放
  • 电子邮件的主题
  • 不使用 LPMetaDataProvider 提取请求(如本 WWDC 2019 262 视频中所述),因此加载速度更快

0。初始化 UIActivityViewController:

将项目设置为 self:

let activityVC = UIActivityViewController(activityItems: [self], applicationActivities: nil)

排除某些不适用的 ActivityType

activityVC.excludedActivityTypes = [.addToReadingList, .assignToContact, .markupAsPDF, .openInIBooks, .saveToCameraRoll]

对于 iPad 设置 popoverPresentationController.sourceView.barButtonItem(这在 iPhone 上被忽略):

activityVC.popoverPresentationController?.sourceView = myButton

出示:

present(activityVC, animated: true, completion: nil)

1。实施所需的 UIActivityItemSource 方法

https://developer.apple.com/documentation/uikit/uiactivityitemsource

您必须根据文档实现占位符方法:

Placeholder objects do not have to contain any real data but should be configured as closely as possible to the actual data object you intend to provide.

func activityViewControllerPlaceholderItem(_ activityViewController: UIActivityViewController) -> Any {
    return ""
}

以及实际数据,将 link 返回给 AirDrop 的应用程序,并为其他所有内容返回文本:

func activityViewController(_ activityViewController: UIActivityViewController, itemForActivityType activityType: UIActivity.ActivityType?) -> Any? {
    
    if activityType == .airDrop {
        return URL(string: "APP_STORE_URL")!
    }
    
    return "Check out the APP_NAME on the App Store: APP_STORE_URL or on the Google Play Store: PLAY_STORE_URL"
}

2。实现主题方法

来自文档:

For activities that support a subject field, returns the subject for the item.

func activityViewController(_ activityViewController: UIActivityViewController, subjectForActivityType activityType: UIActivity.ActivityType?) -> String {
    return "EMAIL_SUBJECT" // e.g. App name
}

3。实施LPLinkMetaData方法

来自文档:

Returns metadata to display in the preview header of the share sheet.

@available(iOS 13.0, *)
func activityViewControllerLinkMetadata(_ activityViewController: UIActivityViewController) -> LPLinkMetadata? {
    let metadata = LPLinkMetadata()
    metadata.title = "APP_NAME"
    return metadata
}