Getopenfilename Default File Path Excel. VBA script to capture file path. API: Call the standard Windows File Open/Save dialog box. Author(s) Ken Getz This can be done by either using the Common Dialog Control in Access 97 or by using. File filter, allow xls, xlsx and xlsm and nothing else.

I'm trying to adopt the Application.GetOpenFilename method, so I can let go of the dependency for comdlg32.ocx, which is not included in Vista. However, I haven't found a way to have GetOpenFilename start with a default filename. Starting in a given directory works (if I call it before calling GetOpenFilename), but I wonder if it's possible to have the dialog box open with a default filename in that directory? That is, when the dialog opens (in the given directory), the default file should be pre-selected. Is that possible? You need to use chdrive and chdir before calling the getopenfilename.

Chdrive('h:') chdir('temp') 'Gustaf' wrote: >I'm trying to adopt the Application.GetOpenFilename method, so I can let go of the dependency for comdlg32.ocx, which is not included in Vista. However, I haven't found a way to have GetOpenFilename start with a default filename. Starting in a given directory works (if I call it before calling GetOpenFilename), but I wonder if it's possible to have the dialog box open with a default filename in that directory? That is, when the dialog opens (in the given directory), the default file should be pre-selected. Is that possible? Thank you, but maybe you didn't read the question in detail.

I'm not asking for default directory, but default filename. Also, when you open default directory, I have found it better to call Private Declare Function SetCurrentDirectory Lib 'kernel32' Alias 'SetCurrentDirectoryA' (ByVal lpPathName As String) As Long rather than ChDir(). It works with UNC (network) paths too. If the answer to my question is 'you can't do that', that would be okay too. Gustaf Joel wrote: >You need to use chdrive and chdir before calling the getopenfilename.

>>>chdrive('h:') chdir('temp') >>'Gustaf' wrote: >>>I'm trying to adopt the Application.GetOpenFilename method, so I >>can let go of the dependency for comdlg32.ocx, which is not >>included in Vista. However, I haven't found a way to have >>GetOpenFilename start with a default filename.

Starting in a given >>directory works (if I call it before calling GetOpenFilename), but >>I wonder if it's possible to have the dialog box open with a >>default filename in that directory? That is, when the dialog opens >>(in the given directory), the default file should be pre-selected.

>>Is that possible? >>>>Gustaf >. I tried various alternatives to 'Text Files (*.txt), *.txt') eg 'Excel Files (*.xls),abc.xls' but they don't 'pre select' the required file. So i guess it can't be done. 'Gustaf' wrote in message news:#. >Thank you, but maybe you didn't read the question in detail.

I'm not >asking for default directory, but default filename. Also, when you open >default directory, I have found it better to call >>Private Declare Function SetCurrentDirectory Lib 'kernel32' Alias >'SetCurrentDirectoryA' (ByVal lpPathName As String) As Long >>rather than ChDir().

It works with UNC (network) paths too. >>If the answer to my question is 'you can't do that', that would be okay >too. >>Gustaf >>Joel wrote: >>You need to use chdrive and chdir before calling the getopenfilename. >>>>>>chdrive('h:') chdir('temp') >>>>'Gustaf' wrote: >>>>>I'm trying to adopt the Application.GetOpenFilename method, so I >>>can let go of the dependency for comdlg32.ocx, which is not >>>included in Vista.

However, I haven't found a way to have >>>GetOpenFilename start with a default filename. Starting in a given >>>directory works (if I call it before calling GetOpenFilename), but >>>I wonder if it's possible to have the dialog box open with a >>>default filename in that directory? That is, when the dialog opens >>>(in the given directory), the default file should be pre-selected. >>>Is that possible? >>>>>>Gustaf >>. You can use the filedialog instead ChDrive ('c: ') ChDir ('c: temp') Set filetoopen = Application.FileDialog(msoFileDialogOpen) filetoopen.InitialFileName = 'book1.xls' filetoopen.Show 'Gustaf' wrote: >Thank you, but maybe you didn't read the question in detail. Getspc Apk Download more. I'm not asking for default directory, but default filename.

Also, when you open default directory, I have found it better to call >>Private Declare Function SetCurrentDirectory Lib 'kernel32' Alias 'SetCurrentDirectoryA' (ByVal lpPathName As String) As Long >>rather than ChDir(). It works with UNC (network) paths too. >>If the answer to my question is 'you can't do that', that would be okay too. >>Gustaf >>Joel wrote: >>You need to use chdrive and chdir before calling the getopenfilename.

>>>>>>chdrive('h:') chdir('temp') >>>>'Gustaf' wrote: >>>>>I'm trying to adopt the Application.GetOpenFilename method, so I >>>can let go of the dependency for comdlg32.ocx, which is not >>>included in Vista. However, I haven't found a way to have >>>GetOpenFilename start with a default filename. Starting in a given >>>directory works (if I call it before calling GetOpenFilename), but >>>I wonder if it's possible to have the dialog box open with a >>>default filename in that directory? That is, when the dialog opens >>>(in the given directory), the default file should be pre-selected. >>>Is that possible?

>>>>>>Gustaf >>>. This is exactly what I needed.

Thank you very much! Gustaf Joel wrote: >You can use the filedialog instead >>ChDrive ('c: ') >ChDir ('c: temp') >Set filetoopen = Application.FileDialog(msoFileDialogOpen) >filetoopen.InitialFileName = 'book1.xls' >filetoopen.Show >>'Gustaf' wrote: >>>Thank you, but maybe you didn't read the question in detail. I'm not asking for default directory, but default filename.

Also, when you open default directory, I have found it better to call >>>>Private Declare Function SetCurrentDirectory Lib 'kernel32' Alias 'SetCurrentDirectoryA' (ByVal lpPathName As String) As Long >>>>rather than ChDir(). It works with UNC (network) paths too. >>>>If the answer to my question is 'you can't do that', that would be okay too. >>>>Gustaf >>>>Joel wrote: >>>You need to use chdrive and chdir before calling the getopenfilename.

>>>>>>>>>chdrive('h:') chdir('temp') >>>>>>'Gustaf' wrote: >>>>>>>I'm trying to adopt the Application.GetOpenFilename method, so I >>>>can let go of the dependency for comdlg32.ocx, which is not >>>>included in Vista. However, I haven't found a way to have >>>>GetOpenFilename start with a default filename.

Starting in a given >>>>directory works (if I call it before calling GetOpenFilename), but >>>>I wonder if it's possible to have the dialog box open with a >>>>default filename in that directory? That is, when the dialog opens >>>>(in the given directory), the default file should be pre-selected.

>>>>Is that possible? >>>>>>>>Gustaf >>>.

Hello, I'm using Excel 2002 on Windows 2000. In some VBA code I need to get the user to open a file from a certain location, or at least get the dialog box to default to that location. I also need to know whether a valid file has been opened and what the file name is. This is what I'm currently using: With Application ' Store the original default file path sOriginalPath =.DefaultFilePath ' Set the default file path to the directory we want.DefaultFilePath = sOpenPath ' Get the name of the file to be opened sOpenFile =.GetOpenFilename('All files (*.*), *.*') ' Restore the original default file path.DefaultFilePath = sOriginalPath End With However, although.DefaultFilePath actually changes, the dialog box ignores it. I've tried creating then closing a new instance of Excel after setting the path; it doesn't make any difference. I've also tried using: Application.Dialogs(xlDial ogOpen).Sh ow sOpenPath to open the file, but that doesn't let me obtain the file name.

The routine below sets the folder name ('C:temp) and then cycles through each file in this folder and adds it to an array and a string if it is a spreadsheet. You could modify this validity test (based on name, type, timestamp etc).

Sub FileNametoExcel() Dim fnam As Variant ' fnam is an array of files returned from GetOpenFileName ' note that fnam is of type boolean if no array is returned. ' That is, if the user clicks ' on cancel in the file open dialog box, fnam is set to FALSE Dim b As Integer 'counter for filname array Dim b1 As Integer 'counter for finding in filename Dim c As Integer 'extention marker ' first open a blank sheet and go to top left ActiveWorkbook.Worksheets. Farquiza, thanks for the replies but your first solution won't return the file name and your second one doesn't let you set the default path. Brettdj, your solution would work but I think PoshDog's solution looks like it's the cleanest. I haven't tried it yet though, I'll have to get my head around FileDialog first. However, after I posted the query I did one last search on-line and managed to find a very easy solution. Here it is: Dim sOpenFile As String ' File to open ' ChDir will change the directory that the 'open file' dialog box ' defaults to, but only if it's on the same drive as the current ' default directory.

If not, you need a ChDrive first. ChDrive (Left(sOpenPath, 3)) ChDir (sOpenPath) ' Get the name of the file to be opened sOpenFile = Application.GetOpenFilenam e('All files (*.*), *.*') I'd tried using ChDir without success, but the key is that you have to use ChDrive first unless your new path is on the same drive as your current path. This should be a PAQ. Here's the answer I've been using in my VBAs it works flawlessly: and the sub I call it from Private Sub Command17_Click() 'Andrew Komasinski 'References: 'X Dim cmdlgOpenFile As New clscommondialog Dim Filename As String 'full file name Const clngFilterIndexAll = 5 cmdlgOpenFile.Filter = 'Excel Spreadsheets (xls) *.xls' cmdlgOpenFile.FilterIndex = clngFilterIndexAll 'this is where the dialog opens cmdlgOpenFile.ShowOpen 'returns your full file name. Filename = cmdlgOpenFile.Filename 'hence no len, no name. If Len(Filename) = 0 Then Exit Sub Dim fs As Object Set fs = CreateObject('Scripting.Fi leSystemOb ject') If fs.FileExists(Filename) Then DoCmd.OpenForm 'Import Wizard' Application.Forms('Import Wizard').MyRealOpen (Filename) 'my Real open was a function I used to open the file so that I could import excel files to access. 'just replace this with whatever openfile you need Else MsgBox 'That file would not open' Exit Sub End If End Sub.