01.
02.If OpenImageFile.ShowDialog = Windows.Forms.DialogResult.OK AndAlso OpenImageFile.FileName <> Nothing Then
03.
04. For Each file As String In OpenImageFile.FileNames
05. Try
06. Dim pic As New PictureBox
07. pic.Image = Image.FromFile(file)
08. Using stream As New IO.MemoryStream
09. pic.Image.Save(stream, System.Drawing.Imaging.ImageFormat.Jpeg)
10. strSQL = "INSERT INTO picture(byte_pic,key_pic,note_pic)" _
11. & "VALUES(@byte_pic,@key_pic,@note_pic)"
12. cmd = New MySqlCommand(strSQL, ConnectionDB)
13. cmd.Parameters.AddWithValue("@byte_pic", stream.GetBuffer())
14. cmd.Parameters.AddWithValue("@key_pic", "images1")
15. cmd.Parameters.AddWithValue("@note_pic", "note")
16. Call open_connection()
17. cmd.ExecuteNonQuery()
18. stream.Close()
19. End Using
20. Catch ex As Exception
21.
22. End Try
23. Next
24. End If
