Filtering single file usig dialog box

[Ryobi]

Hello Everyone,



I am wondering if somebody knows how to fitler a single file name in msaccess (2016) using dialog box. What I tried the following filter:

.Filters.Add "Smaple.accdb", "*.accdb" which works except that also displays all accb files. I only need only show the single file "Smaple.accdb"

Thank you

[davegri]

Try

.Filters.Add "Smaple.accdb"

or if it insists on the second parameter

.Filters.Add "Smaple.accdb", "*.xcqrs"

It's not likely to find that extension.

[Ryobi]

I had already tried .Filters.Add "Sample.accbd" and I got an error because it need the second paramenter. I also tired .Filters "Sample.Accdb", "Sample.Accdb" which give an error on the second paramenter.
When I tried .Filters.Add "Sample.accdb", "*.xcqrs" I get no file listing. I used to used to because work when I a function below (It's the dialog function), but unfornately it does not work with Msaccess 2016. It does not
open file selection option and it does get an message at all, added to fact that is function is for 32 bit and I using 64 bit Msaccess.


strFilter = ahtAddFilterItem(strFilter, "Sample.accdb", "Sample.accdb")
strInputFileName = ahtCommonFileOpenSave( _
Filter:=strFilter, OpenFile:=True, _
DialogTitle:="Please select data file...", _
Flags:=ahtOFN_HIDEREADONLY)

[isladogs]

The following code works in 32-bit & 64-bit Access

 Set options for the dialog box.    Dim F As FileDialog
    Set F = Application.FileDialog(msoFileDialogFilePicker)
    F.title = "Locate the JSON file folder and click on 'Open' to select it" 'modify as appropriate
    
' Clear out the current filters, and add our own
    F.Filters.Clear
    F.Filters.Add "JSON files", "*.json"  'modify as appropriate
    
' Set the start folder. Open in default file folder if blank
    F.InitialFileName = Nz(Application.CurrentProject.Path & "\Files\", "C:\") 'modify as appropriate
    
' Call the Open dialog procedure.
    F.Show


' Return the path and file name.
    strFilePath = F.SelectedItems(1)
    
    Me.FileName = Mid(strFilePath, InStrRev(strFilePath, "\") + 1)
    strFile = Me.FileName
    Me.FolderPath = Left(strFilePath, Len(strFilePath) - Len(strFile) - 1)
    
    strFilename = Left(strFile, InStr(strFile, ".") - 1)

However that doesn't solve your issue of filtering for an individual file
I don't believe its possible to do so using a File...Open dialog
By definition it filters for the file type you specify and shows all the files of that type
If you specify a nonsense file type, you will get no results

However I don't see the point of doing this anyway.
If you know the file that you want to open, then scrap the file open dialog box
Just open the file direct from the button click event

[davegri]

I agree with Colin. If you need to verify that a specific file exists before opening it, check out the DIR function.
https://support.office.com/en-us/art...7-41e4513bba2e

[Ryobi]

ridder52,

I have not tried what you wrote on the post, but will this do that later in the day. Regarding opening the the file directly, it may be possible that there my be other access files, for example the program that I have has
two access file 1. program file Prg.accdb (This file a querys, forms and reports) and 2.) data file Prgdata.accdb (this file only has tables which are linked to Prg.accdb). By using this method I can alway modify the the program without affect the data, other wise you will comprise the data every time you modify the program (of course you could a rely on backup file and start again the modifications). I have a from in the program where I used the dialogbox to get file and linke (Not open) the tables from the datafile. I hope this means sense to you.

[isladogs]

ridder52,

I have not tried what you wrote on the post, but will this do that later in the day. Regarding opening the the file directly, it may be possible that there my be other access files, for example the program that I have has
two access file 1. program file Prg.accdb (This file a querys, forms and reports) and 2.) data file Prgdata.accdb (this file only has tables which are linked to Prg.accdb). By using this method I can alway modify the the program without affect the data, other wise you will comprise the data every time you modify the program (of course you could a rely on backup file and start again the modifications). I have a from in the program where I used the dialogbox to get file and linke (Not open) the tables from the datafile. I hope this means sense to you.

Not really.
You have a split database with Access FE and BE.
Beyond that I can't understand what you're saying.
Where does 'sample.accdb' come into this?

[Ryobi]

I am sorry, but I change the file names when I explained the problem. I used Prg.accdb as the name of the program (front end) and Prgdata.accb as the data (backend). I was substituting Prgdata.accb with Sample.accdb (the data file). I could not find a way to show only that file so I did a work around. If they select any file other the one that the program needs they will received an error message. I have the code below.

For Each varFile In .SelectedItems

strInputFileName = Dir(varFile)
If strInputFileName <> "Sample.accdb" Then
MsgBox strInputFileName & " is not a valid file data file, please select Sample.accb", , "** No valid data File **"
Exit Sub
End If
Next