Mam następujący kod do dzielenia obrazu na Instagram w Swift 3 iOS 10.1:akcji iOS na Instagram: Plik nie może zostać zapisany, ponieważ określony typ URL nie jest obsługiwana
func shareOnInstagram(_ photo: UIImage, text: String?) {
let instagramUrl = URL(string: "instagram://app")!
if UIApplication.shared.canOpenURL(instagramUrl) {
let imageData = UIImageJPEGRepresentation(photo, 1.0)!
let captionString = text ?? ""
let writePath = URL(string: NSTemporaryDirectory())!.appendingPathComponent("instagram.igo")
do {
try imageData.write(to: writePath)
let documentsInteractionsController = UIDocumentInteractionController(url: writePath)
documentsInteractionsController.delegate = self
documentsInteractionsController.uti = "com.instagram.exlusivegram"
documentsInteractionsController.annotation = ["InstagramCaption": captionString]
documentsInteractionsController.presentOpenInMenu(from: CGRect.zero, in: self.view, animated: true)
}catch {
return
}
}else {
let alertController = UIAlertController(title: "Install Instagram", message: "You need Instagram app installed to use this feature. Do you want to install it now?", preferredStyle: .alert)
let installAction = UIAlertAction(title: "Install", style: .default, handler: { (action) in
//redirect to instagram
})
alertController.addAction(installAction)
let laterAction = UIAlertAction(title: "Later", style: .cancel, handler: nil)
alertController.addAction(laterAction)
present(alertController, animated: true, completion: nil)
}
}
An u linia try imageData.write(to: writePath)
, to rzuca błąd w następujący sposób:
Error Domain=NSCocoaErrorDomain Code=518 "The file couldn’t be saved because the specified URL type isn’t supported." UserInfo={NSURL=/private/var/mobile/Containers/Data/Application/5B80A983-5571-44A5-80D7-6A7B065800B5/tmp/instagram.igo}
Czy ktoś może mi pomóc z tym?