ios - 在 iOS Swift 上发布到 Instagram 屏幕

Question

我正在努力简化从我的应用程序到 Instagram 的分享。我想要的是进入下面屏幕截图中描述的屏幕。我试过 instagram-stories://share deeplink 并且我通读了所有这些文档:https://developers.facebook.com/docs/instagram/sharing-to-stories/

但是，无论我做什么，当 url scheme Action 触发时，它会直接将图像分享到故事中。我在这里缺少什么？

这是我的代码摘录:

           if let image = image {
                guard let urlScheme = URL(string: "instagram-stories://share"),
                    let imageData = image.pngData() else {
                    return
                }

                if UIApplication.shared.canOpenURL(urlScheme) {
                    let pasterboardItems = [["com.instagram.sharedSticker.backgroundImage": imageData]]
                    let pasterboardOptions = [UIPasteboard.OptionsKey.expirationDate: Date().addingTimeInterval(60*5)]

                    UIPasteboard.general.setItems(pasterboardItems, options: pasterboardOptions)

                    UIApplication.shared.open(urlScheme, options: [:], completionHandler: nil)
                }
            }

Answer 1

您需要做的是使用以下网址打开 Instagram 应用程序:
instagram://library?LocalIdentifier= 并作为参数传递 PHAsset.localIdentifier。
由于某种原因，这个钩子(Hook)没有在文档中的任何地方列出🤷‍♂️

但是为了接收您的图像/视频的本地标识符，您必须先将图像/视频保存到用户的照片库中。所以最终的代码看起来像这样

let videoFileUrl: URL = URL(fileURLWithPath: "path/to/my/video")!
var localId: String?
PHPhotoLibrary.shared().performChanges({
    let request = PHAssetChangeRequest.creationRequestForAssetFromVideo(atFileURL: videoFileUrl)
    localId = request?.placeholderForCreatedAsset?.localIdentifier
}, completionHandler: { success, error in
    // completion handler is called on an arbitrary thread
    // but since you (most likely) will perform some UI stuff
    // you better move everything to the main thread.
    DispatchQueue.main.async {
        guard error == nil else {
            // handle error
            return
        }
        guard let localId = localId else {
            // highly unlikely that it'll be nil,
            // but you should handle this error just in case
            return
        }

        let url = URL(string: "instagram://library?LocalIdentifier=\(localId)")!
        guard UIApplication.shared.canOpenURL(url) else {
            // handle this error
            return
        }
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
})