cannot get GetOpenFilename to work

[scotgotchi]

Could someone please take a look at this bit of code and tell me what's
wrong with it - I'm stumped, I've tried various things and just cannot
get it to work. I'm using Windows XP and Office 2003 .
Ultimately, I'll want it to default to a network drive, but I can't
even get it to work when starting on C:\. I get a runtime error 1004
which I believe is "file not found".

Sub myOpenFileDialog()
'
' myOpenFileDialog Macro
'
myFilt = "Text Files (*.txt), *.txt" & _
"Comma Separated Values (*.csv), *.csv" & _
"Excel Workbooks (*.xls), *.xls" & _
"All Files (*.*), *.*"
myFilterIndex = 4
myTitle = "Select Data File to Open"

ChDrive "C:\"
ChDir "C:\"
' Get the file name
myDataFileName = Application.GetOpenFilename(FileFilter:=myFilt, _
FilterIndex:=myFilterIndex, Title:=myTitle)

' Exit if user cancels dialog box
If myDataFileName = False Then
MsgBox "No file Selected"
Exit Sub
End If

' Show myDataFileName Result
MsgBox "Data Filename = " & myDataFileName

End Sub

[Ardus Petus]

FilterIndex needs comma-separated values:
myFilt = "Text Files (*.txt), *.txt," & _
"Comma Separated Values (*.csv), *.csv," & _
"Excel Workbooks (*.xls), *.xls," & _
"All Files (*.*), *.*"

HTH
--
AP

"scotgotchi" <[email protected]> a écrit dans le message de news:
[email protected]...
> Could someone please take a look at this bit of code and tell me what's
> wrong with it - I'm stumped, I've tried various things and just cannot
> get it to work. I'm using Windows XP and Office 2003 .
> Ultimately, I'll want it to default to a network drive, but I can't
> even get it to work when starting on C:\. I get a runtime error 1004
> which I believe is "file not found".
>
> Sub myOpenFileDialog()
> '
> ' myOpenFileDialog Macro
> '
> myFilt = "Text Files (*.txt), *.txt" & _
> "Comma Separated Values (*.csv), *.csv" & _
> "Excel Workbooks (*.xls), *.xls" & _
> "All Files (*.*), *.*"
> myFilterIndex = 4
> myTitle = "Select Data File to Open"
>
> ChDrive "C:\"
> ChDir "C:\"
> ' Get the file name
> myDataFileName = Application.GetOpenFilename(FileFilter:=myFilt, _
> FilterIndex:=myFilterIndex, Title:=myTitle)
>
> ' Exit if user cancels dialog box
> If myDataFileName = False Then
> MsgBox "No file Selected"
> Exit Sub
> End If
>
> ' Show myDataFileName Result
> MsgBox "Data Filename = " & myDataFileName
>
> End Sub
>

[scotgotchi]

Thanks very much Ardus.
So Simple!