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ACC: FileCopy Statement May Not Copy Open Files

This article was previously published under Q172711
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This article was written about products for which Microsoft no longer offers support. Therefore, this article is offered "as is" and will no longer be updated.
Advanced: Requires expert coding, interoperability, and multiuser skills.

SYMPTOMS
When you programmatically copy a file with the FileCopy statement in VisualBasic for Applications, you may receive the following error message:
Run-time error '70'

Permission denied

This article assumes that you are familiar with Visual Basic forApplications and with creating Microsoft Access applications using theprogramming tools provided with Microsoft Access. For more informationabout Visual Basic for Applications, please refer to your version of the"Building Applications with Microsoft Access" manual.
CAUSE
The file is currently open, which prevents the FileCopy statement fromcopying the file.
RESOLUTION
Instead of using the FileCopy statement, use one of the following methodsto programmatically copy the file.

WARNING: The following functions enable you to copy an open file. If thesource file is changed while the copy operation is in process, thedestination file may be incomplete or may become corrupted.

Method 1 - Calling the CopyFile() function from the Windows API

One method to programmatically copy a file is to call the CopyFile()function from the Microsoft Windows API. To call the CopyFile() functionfrom the Microsoft Windows API, follow these steps:
  1. Complete steps 1 through 4 from the "Steps to Reproduce Behavior" section later in this article.
  2. Create a module and type the following lines in the Declarations section:
          Option Explicit						
          Declare Function apiCopyFile Lib "kernel32" Alias "CopyFileA" _      (ByVal lpExistingFileName As String, _      ByVal lpNewFileName As String, _      ByVal bFailIfExists As Long) As Long						
  3. Type the following procedure:
          Sub CopyFile(SourceFile As String, DestFile As String)      '---------------------------------------------------------------      ' PURPOSE: Copy a file on disk from one location to another.      ' ACCEPTS: The name of the source file and destination file.      ' RETURNS: Nothing      '---------------------------------------------------------------        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						
  4. To test this procedure, type the following line in the Debug window, and then press ENTER:

    CopyFile "<path to Northwind.mdb>", "C:\Northwind.mdb"

    Note that Northwind.mdb is copied to the root folder of drive C, even though it is currently open in another instance of Microsoft Access.

Method 2 - Calling the MS-DOS Copy Command

Another method to programmatically copy a file is to call the MS-DOS Copycommand from a Shell() function in Visual Basic for Applications. To callthe MS-DOS Copy command, follow these steps:
  1. Complete steps 1 through 4 from the "Steps to Reproduce Behavior" section later in this article.
  2. Create a module and type the following line in the Declarations section if it is not already there:
          Option Explicit						
  3. If you are using Microsoft Windows 95, type the following procedure:
          Sub CopyFile(SourceFile As String, DestFile As String)      '---------------------------------------------------------------      ' PURPOSE: Copy a file on disk from one location to another.      ' ACCEPTS: The name of the source file and destination file.      ' RETURNS: Nothing      '---------------------------------------------------------------         Dim CopyString As String         If Dir(SourceFile) = "" Then            MsgBox Chr(34) & SourceFile & Chr(34) & _               " is not a valid file name."         Else            SourceFile = Chr(34) & SourceFile & Chr(34)            DestFile = Chr(34) & DestFile & Chr(34)            CopyString = "COMMAND.COM /C COPY " & SourceFile & _               " " & DestFile            Call Shell(CopyString, 0)         End If      End Sub   If you are using Microsoft Windows NT, use the same procedure, but   change the line      CopyString = "COMMAND.COM /C COPY " & SourceFile & _   to:      CopyString = "CMD.EXE /C COPY " & SourceFile & _						
  4. To test this procedure, type the following line in the Debug window, and then press ENTER:

    CopyFile "<path to Northwind.mdb>", "C:\Northwind.mdb"

    Note that Northwind.mdb is copied to the root folder of drive C, even though it is currently open in another instance of Microsoft Access.
STATUS
This behavior is by design.
MORE INFORMATION

Steps to Reproduce Behavior


  1. Start Microsoft Access.
  2. Open the sample database Northwind.mdb.
  3. Start a new instance of Microsoft Access.
  4. Create a new blank database.
  5. Create a module and type the following line in the Declarations section if it is not already there:
          Option Explicit						
  6. Type the following procedure:
          Sub CopyFile(SourceFile As String, DestFile As String)         FileCopy SourceFile, DestFile      End Sub						
  7. To test this procedure, type the following line in the Debug window, and then press ENTER:

    CopyFile "<path to Northwind.mdb>", "C:\Northwind.mdb"

    Note that you receive the error listed in the "Symptoms" section.
REFERENCES
For more information about the FileCopy statement, search the Help Indexfor "FileCopy statement."
file copy
Properties

Article ID: 172711 - Last Review: 01/20/2007 02:15:30 - Revision: 3.1

  • Microsoft Access 95 Standard Edition
  • Microsoft Access 97 Standard Edition
  • kbprb kbprogramming KB172711
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