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

If you would like to do the same without using the api:

Function SharedFilecopy(ByVal SourcePath As String, ByVal DestinationPath As String)

Dim FF1 As Long, FF2 As Long
Dim Index As Long
Dim FileLength As Long
Dim LeftOver As Long
Dim NumBlocks As Long
Dim filedata As String
Dim ErrCount As Long
On Error GoTo ErrorCopy
'-------------
'Copy the file
'-------------
Const BlockSize = 32767
FF1 = FreeFile
Open SourcePath$ For Binary Access Read As #FF1
FF2 = FreeFile
Open DestinationPath For Output As #FF2
Close #FF2

Open DestinationPath For Binary As #FF2

Lock #FF1: Lock #FF2

FileLength = LOF(FF1)
NumBlocks = FileLength \ BlockSize
LeftOver = FileLength Mod BlockSize

filedata = String$(LeftOver, 32)

Get #FF1, , filedata
Put #FF2, , filedata
filedata = ""
filedata = String$(BlockSize, 32)

For Index = 1 To NumBlocks
    Get #FF1, , filedata
    Put #FF2, , filedata
Next Index
Unlock #FF1: Unlock #FF2
SharedFilecopy = True

exitcopy:

Close #FF1, #FF2

Exit Function

ErrorCopy: ErrCount = ErrCount + 1

If ErrCount > 2000 Then

SharedFilecopy = False

Resume exitcopy

Else

Resume

End If

End Function