Advertisement
Guest User

Untitled

a guest
Jul 17th, 2019
88
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
VB.NET 0.48 KB | None | 0 0
  1. Private Sub SurroundingSub()
  2.     Dim fileMetadata = New File() With {
  3.         .Name = "photo.jpg"
  4.     }
  5.     Dim request As FilesResource.CreateMediaUpload
  6.  
  7.     Using stream = New System.IO.FileStream("files/photo.jpg", System.IO.FileMode.Open)
  8.         request = driveService.Files.Create(fileMetadata, stream, "image/jpeg")
  9.         request.Fields = "id"
  10.         request.Upload()
  11.     End Using
  12.  
  13.     Dim file = request.ResponseBody
  14.     Console.WriteLine("File ID: " & file.Id)
  15. End Sub
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement