I'm using: Application.GetOpenFilename(MultiSelect:=False)
so my client can find a file.
However, I've already saved the top level directory
And do not want my client to re-find that directory,
Instead, I'd like GetOpenFinename() to start in this top level directory.
GetOpenFileName() starts in what ever directory was last looked at.
How can I change this directory?
I have searched the forum for this problem, and haven't been able to find it. Is there a way, when using the GetOpenFilename method, of forcing the Open control window to appear on TOP of any windows currently open? It is always opening below whatever I have open, but doesn't appear on the taskbar. In order to find it, I always have to do Alt-Tab. This program is going to be used by multiple users, so needs to be as straight-forward as possible.
This is the code currently being used:
Dim mstrFileName As String
Private xlExcel As Excel.Application, objBooks As Object, xlWorkBook As Excel.Workbook
mstrFileName = xlExcel.GetOpenFilename
Is there a method of using the GetOpenFileName FilterIndex parameter to retrieve the last file in the directory?
I'm currently using the below to retrieve a specific file but would like to bypass the dialog box and just go out and get the last file put into the folder.
ChDir "P:\IntrDept\Field Reports\Jims\Important Data" fileToOpen = Application _ .GetOpenFilename("CSV Files (*.csv), *.csv") If fileToOpen <> False Then MsgBox "Open " & fileToOpen Workbooks.Open fileToOpen, False, True
Apparently, I am doing something wrong.
I am attempting to let the user select multiple file names with the GetOpenFileName method.
Dim vFile As Variant, vFile2 As Variant vFile = Application.GetOpenFileName(FileFilter:="Excel Files (*.xls),*.xls",MultiSelect:=True) 'So far so good For Each vFile2 In vFile 'generates an error I = UBound (vFile) 'Also generates an error
What am I doing wrong? This seems easy, but.
I want to open a path with a command button.
I always use "application.getopenfilename" to open files but the I get the path+filename but I only want the path. and not the filename.
I know it's possible to open a file with application.getopenfile and extract the path out of it but I want the same screen as the applicatio.getopenfilename but without the files, only the folders.
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