I'm using the code below to display a dialog box for the user to select a
file. I'm then checking to see if the file is opened. However, it appears
the code is running faster than the file opens and I and getting the error
message that the file was not opened when in fact it is.

Is there a way to do slow down the code until the file is opened without
coding a hard pause of X number of seconds?

Any help or suggestion would be greatly appreciated. Thanks!

FName = Application.GetOpenFilename

If FName <> False Then
Workbooks.Open (FName)
End If

If Workbooks.Count <> Wcount + 1 Then
MsgBox "The file was not opened."
Sheets("Macros").Activate
End If

code continues here.......

--
JT

Hi,

Your code will pause at

> FName = Application.GetOpenFilename

but a general method for checking if a workbook is open or not is

Sub IsOpen()
On Error Resume Next

Set wBook = Workbooks(FName)
If wBook Is Nothing Then 'Not open
MsgBox "Workbook is not open", vbCritical
Set wBook = Nothing
On Error GoTo 0
Else 'It is open
MsgBox "Yes it is open", vbInformation
Set wBook = Nothing
On Error GoTo 0
End If
End Sub

Mike

"JT" wrote:

> I'm using the code below to display a dialog box for the user to select a
> file. I'm then checking to see if the file is opened. However, it appears
> the code is running faster than the file opens and I and getting the error
> message that the file was not opened when in fact it is.
>
> Is there a way to do slow down the code until the file is opened without
> coding a hard pause of X number of seconds?
>
> Any help or suggestion would be greatly appreciated. Thanks!
>
> FName = Application.GetOpenFilename
>
> If FName <> False Then
> Workbooks.Open (FName)
> End If
>
> If Workbooks.Count <> Wcount + 1 Then
> MsgBox "The file was not opened."
> Sheets("Macros").Activate
> End If
>
> code continues here.......
>
> --
> JT