Question

I have this segment of code:

    Dim myStream As Stream = Nothing
    Dim openFileDialog1 As New OpenFileDialog()

    openFileDialog1.InitialDirectory = "C:\Users\Desktop\Sample Pictures"
    openFileDialog1.Filter = "Pictures|*.jpg|Text|*.txt"

    If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then
            myStream = openFileDialog1.OpenFile()
            If (myStream IsNot Nothing) Then

              ' Insert code to read the stream here.

            End If
        End Sub
    End Class

I need to copy the file in myStream to some destination Folder.
Any idea how I can implement it?

Answer 1

Supposing the myDestinationDir is the path where you want to copy your file then

If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then 
    ' Extract the filename part from the full filename returned by openDialog.'
    Dim selectedFile As String = Path.GetFileName(openFileDialog1.FileName)
    ' Append to the destinationDir the filename extracted'
    Dim destFile As String = Path.Combine(myDestinationDir, selectedFile)
    ' Copy with overwrite (if overwrite is not desidered, use the overload with False as 3 arg'
    System.IO.File.Copy(openFileDialog1.FileName, destFile)
End Sub 

This will copy the selected file to the destination folder overwriting an existing file with the same name.