在iOS Swift上发布到Instagram屏幕 [英] Post To Instagram screen on iOS Swift

问题描述

我正在努力让从我的应用程序到Instagram的分享变得容易。我想要的是转到下面屏幕截图上所描绘的屏幕。我试过Instagram-Stories：//分享深度链接，我已经阅读了所有这些文档：https://developers.facebook.com/docs/instagram/sharing-to-stories/

然而，无论我做什么，当url方案操作触发时，它直接将图像分享到故事中。我在这里错过了什么？

以下是我的代码摘录：

           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)
                }
            }

推荐答案

您需要做的是使用以下URL打开Instagram应用程序：
instagram://library?LocalIdentifier=并作为参数PHAsset.localIdentifier传递。
由于某种原因，此挂钩未在文档🤷‍♂️中的任何位置列出

但为了接收您的图像/视频的本地标识，您必须首先将图像/视频保存到用户的图片库。 因此，最终代码将如下所示

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)
    }
})

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