【发布时间】:2017-05-17 04:59:25
【问题描述】:
func shareVideoToInstagram() {
let strURL = "http://mobmp4.org/files/data/2480/Tutak%20Tutak%20Tutiya%20Title%20Song%20-%20Remix%20-%20Drunx%20-%20Mp4.mp4"
let caption = "Some Preloaded Caption"
let captionStr = caption.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)! as String
let videoURL = URL(fileURLWithPath: strURL, isDirectory: false)
let library = ALAssetsLibrary()
library.writeVideoAtPath(toSavedPhotosAlbum: videoURL) { (newURL, error) in
if let instagramURL = NSURL(string: "instagram://library?AssetPath=\(videoURL.absoluteString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!)&InstagramCaption=\(captionStr)") {
print(instagramURL)
if UIApplication.shared.canOpenURL(instagramURL as URL) {
UIApplication.shared.openURL(instagramURL as URL)
}
} else {
print("NO")
}
}
}
我得到这样的 instagramURL:
instagram://library?AssetPath=file:%2F%2F%2Fhttp:%2Fmobmp4.org%2Ffiles%2Fdata%2F2480%2FTutak%252520Tutak%252520Tutiya%252520Title%252520Song%252520-%252520Remix%252520-%25252 %252520-%252520Mp4.mp4&InstagramCaption=Some%20Preloaded%20Caption
我成功地打开了 URL,但我找不到我想在 instagram 上分享的视频。
【问题讨论】:
-
减少代码缩进,但使缩进统一,并将响应格式化为报价
-
现在我成功在 Instagram 上分享视频了。
-
感谢您让我们知道,但如果您添加一个解释您是如何做到的答案会很好?否则我们知道你成功与否对我们没有任何好处,对吧? :\
-
你能告诉我们或在这里分享代码,以便我们也可以看到它是如何完成的吗?
标签: ios swift instagram sharing