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Date modified info for all files in a directory-MS Access VBA

This is kind of an add on question related to a question that I posted a while back (See Link).

I was faced with a situation where I needed to copy the Date Modified information from files and folders located in a specified directory, to a database table.  boag2000 provided a great example that helpd solve the issue I was dealing with.  Now, I would like to know if the code can be expanded upon in order to copy the Date Modified information from all files located in a specific directory, including subfolders, to a database table.

Any input is appreciated.
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1 Solution
You can create table with file name  and Date Modified fields (let table will be tblFiles, fields: FName and FMod), open it and add records in loop:
Dim rst as RecordSet
Set rst = CurrentDB.OpenRecordset ("tblFiles")
For Each fil In fldr.Files
        Debug.Print fil.Path, fil.DateCreated
        rst!Fname = fil.Path
        rst!Fmod = fil.DateCreated

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giswestAuthor Commented:

In your example, don't I need to define the directory where the folders reside?  I defined the fldr variable (SEE EXAMPLE BELOW), but I recieve a Compile Error: Invalid Qualifier.  Based on the error, I don't that it likes the way I defined the variable.


Dim fldr as String

fldr = "C:\SomeFolder\"
Jeffrey CoachmanMIS LiasonCommented:
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Check this combination of two solutions
giswestAuthor Commented:

After looking at your sample database, I modified my code to the following:

Dim rst As Recordset
Dim attachmentPathRemote As String
Dim sourceFolder As Folder
Dim sourceSubFolder As Folder
Dim fileItem As File
Dim fso As Object

DoCmd.SetWarnings False
DoCmd.RunSQL "DELETE * FROM Attachment_File_List_Remote"
DoCmd.SetWarnings True

attachmentPathRemote = Form_AttachmentsBeta.DatabaseLocationLoad.Value & "Attachments\"

Set fso = CreateObject("Scripting.FileSystemObject")
Set rst = CurrentDb.OpenRecordset("Attachment_File_List_Remote")

    Set sourceFolder = fso.GetFolder(attachmentPathRemote)
    On Error Resume Next
    For Each sourceSubFolder In sourceFolder.SubFolders
        For Each fileItem In sourceSubFolder.Files
            rst!File_Name = fileItem.Name
            'rst!Folder_Name = fileItem.ParentFolder.Name
    Set fileItem = Nothing
    Set sourceFolder = Nothing
    Set sourceSubFolder = Nothing
    Set fso = Nothing
Set rst = Nothing

This produces the result I'm looking for, which is to copy the name and date modified of all files contained the directory subfolders into a specified table.

However, I would also like to copy the name of the Parent Folder (Subfolder) that contains the files.  I've tried adding the following line:

rst!Folder_Name = fileItem.ParentFolder.Name

However, this only returns a single record for each subfolder that contains a file.

For example, my current directory contains 8 folders, with only 5 of the 8 folders containing files.  Without adding the "rst!Folder_Name = fileItem.ParentFolder.Name", the names of all files (13 total) contained in the directory folders are copied to the designated table.  With the "rst!Folder_Name = fileItem.ParentFolder.Name", only the names of the folders containing a file (5 folders) and the name of the first file listed in the folders are copied to the designated table.

Any ideas/work arounds regarding this issue?
giswestAuthor Commented:

Your sample worked perfectly when I ran it using your table (tblFileList) as the destination table.  When I defined my table (Attachment_File_List_Remote) as the destination table, I ran into the same issue that I previously described, where only the name and date modified information of the folders containing a file were copied.  I copied your table to my database, renamed it (Attachment_File_List_Remote), and re-ran the code.  Everything worked fine.  I re-ran the code I posted earlier, and it also worked fine.  There must have been an issue with my original table, or I must have had some sort of filter on it, which caused the odd behaviour.  Anyhow, I really appreciate the time and you put into solving this issue.

Have a good day.
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