How can I copy an open file using VB6?

Question

I have a legacy VB6 application that uploads file attachments to a database BLOB field. It works fine unless a user has the file open.

I tried creating a copy of th

Answer 1

Answering my own question:

Based on this article, the answer that worked for me is described below.

1 - Add this declaration to the VB file:

Declare Function apiCopyFile Lib "kernel32" Alias "CopyFileA" _
      (ByVal lpExistingFileName As String, _
      ByVal lpNewFileName As String, _
      ByVal bFailIfExists As Long) As Long

2 - Create a little wrapper for that function, like so:

Sub CopyFileEvenIfOpen(SourceFile As String, DestFile As String)
  Dim Result As Long
   If Dir(SourceFile) = "" Then
     MsgBox Chr(34) & SourceFile & Chr(34) & " is not valid file name."
   Else
     Result = apiCopyFile(SourceFile, DestFile, False)
   End If
End Sub

3 - Replace my previous call to FileCopy with this:

CopyFileEvenIfOpen sourceFile, tempFile