从iOS将视频分享到Instagram feed

Question

我一直在尝试为我们的应用程序创建共享体验，其中Instagram启动时提供了以下两个选项：

Facebook有一个非常精简的文档。我使用UIDocumentInteractionController尝试了所有可能的排列。我尝试使用作为uti com.instagram.photo和com.instagram.video与ig扩展，但我不断收到标准共享酥料饼的，而不是直接启动的Instagram。也尝试过com.instagram.exclusivegram，igo但这似乎应该触发标准弹出窗口。

最新代码：

func shareVideo(_ filePath: String) {
  let url = URL(fileURLWithPath: filePath)
  if(hasInstagram()){
    let newURL = url.deletingPathExtension().appendingPathExtension("ig")
    do {
      try FileManager.default.moveItem(at: url, to: newURL)
    } catch { print(error) }

    let dic = UIDocumentInteractionController(url: newURL)
    dic.uti = "com.instagram.photo"
    dic.presentOpenInMenu(from: self.view.frame, in: self.view, animated: true)
  }
}

Answer 1

进入上图所示屏幕的唯一方法是首先将视频保存在库中，然后使用instagram://library传递资产的未记录挂钩localIdentifier。不要忘记instagram在info.plist.

if UIApplication.shared.canOpenURL("instagram://app") { // has Instagram
    let url = URL(string: "instagram://library?LocalIdentifier=" + videoLocalIdentifier)

    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: [:], completionHandler:nil)
    }
}

Answer 2

尝试这个 ：-

我目前正在通过以下方式共享我上次保存的视频：

    let fetchOptions = PHFetchOptions()
    fetchOptions.sortDescriptors = [NSSortDescriptor(key: "creationDate", ascending: false)]
    let fetchResult = PHAsset.fetchAssets(with: .video, options: fetchOptions)
    if let lastAsset = fetchResult.firstObject {
        let localIdentifier = lastAsset.localIdentifier
        let u = "instagram://library?LocalIdentifier=" + localIdentifier
        let url = NSURL(string: u)!
        if UIApplication.shared.canOpenURL(url as URL) {
            UIApplication.shared.open(URL(string: u)!, options: [:], completionHandler: nil)
        } else {

            let urlStr = "https://itunes.apple.com/in/app/instagram/id389801252?mt=8"
            if #available(iOS 10.0, *) {
                UIApplication.shared.open(URL(string: urlStr)!, options: [:], completionHandler: nil)

            } else {
                UIApplication.shared.openURL(URL(string: urlStr)!)
            }
        }

    }