15. Dim connection As New SqlConnection("connection string here")
16. Dim command As New SqlCommand("UPDATE MyTable SET Picture = @Picture WHERE ID = 1", connection)
18. Using picture As Image = Image.FromFile("file path here")
19. Using stream As New IO.MemoryStream
20. picture.Save(stream, Imaging.ImageFormat.Jpeg)
21. command.Parameters.Add("@Picture", SqlDbType.VarBinary).Value = stream.GetBuffer()
22. End Using
23. End Using
That works. If you have an existing row to update. If you want to insert a new one, the code changes a bit:
16. Dim command As New SqlCommand("INSERT INTO MyTable (Picture) VALUES (@Picture); SELECT @@IDENTITY AS 'Identity'", connection)
17. Dim ThisPictureID As Integer
26. ThisPictureID = CInt(command.ExecuteScalar())
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