我正在尝试简化从我的应用程序到Instagram的共享。我想要的是进入下面的屏幕截图所示的屏幕。我已经尝试过instagram-stories:// share deeplink,并且已经阅读了所有这些文档:https://developers.facebook.com/docs/instagram/sharing-to-stories/
但是,无论我做什么,当url方案动作触发时,它都会直接将图像共享到故事中。我在这里想念什么?
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)
}
}
答案 0 :(得分:1)
image
是“ png” LSApplicationQueriesSchemes
)答案 1 :(得分:1)
您需要做的是使用以下URL打开Instagram应用:
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?♂️
但是为了接收图像/视频的本地标识符,您必须首先将图像/视频保存到用户的照片库。 所以最终代码看起来像这样
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)
}
})