filedialog compile err..using lib 11..

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nycdon - 31 Mar 2005 12:09 GMT

i'm still getting compile error on dim stmt below, even after setting
reference to MS Lib 11..any ideas? (access 2003)

Dim dlgOpen As FileDialog

Set dlgOpen = Application.FileDialog( _
FileDialogType:=msoFileDialogOpen)

With dlgOpen
.AllowMultiSelect = True
.Show
End With

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Terry - 31 Mar 2005 14:53 GMT

Go with Douglas' solution. I have had the API version running nicely for a
fair while. Recently I thought the FileDialog would be a neater solution as
it also included a folder browse - it was a nice solution BUT, the problem
with references was an utter pain. I have returned to the API call. If you
have Access Developer solutions for both the FileDialog and the Folder Browse
are in the Code Librarian or use Douglas' hLink
Terry

> i'm still getting compile error on dim stmt below, even after setting
> reference to MS Lib 11..any ideas? (access 2003)
[quoted text clipped - 8 lines]
> .Show
> End With

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Dirk Goldgar - 31 Mar 2005 21:55 GMT

> i'm still getting compile error on dim stmt below, even after setting
> reference to MS Lib 11..any ideas? (access 2003)
>
> Dim dlgOpen As FileDialog

*Which* "MS Lib 11"? "Microsoft Office 11.0 Object Library" would be
what you need; a reference to "Microsoft *Access* 11.0 Object Library"
(emphasis mine) isn't enough. But I agree with Doug Steele and
others -- calling the API directly is really the way to go.

Signature

Dirk Goldgar, MS Access MVP
www.datagnostics.com

(please reply to the newsgroup)

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Dirk Goldgar - 31 Mar 2005 22:06 GMT

>> i'm still getting compile error on dim stmt below, even after setting
>> reference to MS Lib 11..any ideas? (access 2003)
[quoted text clipped - 6 lines]
> and
> others -- calling the API directly is really the way to go.

I just saw from a separate message of yours that you *are* referencing
the Office library. So I don't know why you would be getting an error
on the Dim statement. When I test this in Access 2002, though, I get an
error on the line

> Set dlgOpen = Application.FileDialog( _
> FileDialogType:=msoFileDialogOpen)

because the example in the help file is just plain wrong, and the
argument should be specified as:

Set dlgOpen = Application.FileDialog( _
DialogType:=msoFileDialogOpen)

or even just

Set dlgOpen = Application.FileDialog(msoFileDialogOpen)

Signature

Dirk Goldgar, MS Access MVP
www.datagnostics.com

(please reply to the newsgroup)

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