Create D/B records that are entered from files saved in directories


I have files which are delivered through DropBox to my file server, what I'd like to do it when a file is 'dropped', something scans the directories, finds the latest saved file, enters a record in the DB which points to this file.

Is this possible? The naming convention of all the files is the same so the format won't change, the directory it resides in though is the differentiator.

So, the filename is like this:


the length of each part won't change, so the first part IMG00001 will always be IMG003432 for example, the format will never change.

The directories are similar:


the second and third part of the structure will never change from a 2 digit/5 digit structure, the last part could be 2-4 alphanumerics.

I know this is probably impossible but thought I'd ask the question anyway!!

Any help would be much appreciated.



Who is Participating?
nrbreenConnect With a Mentor Commented:
The following VBscript will scan all folders under the nominated TopFolder,
locate any JPG file, and add the folderpath, filename, and added-date-time  to a DB table.
It can be run either from the server, or from another machine that can access the folders.
An ODBC DSN must be setup to point to your DB, and you must create an appropriate table.
The script can be run at regular intervals via Windows Scheduler.
You can test it by running from a command prompt, after changing highlighted values to suit your setup.

' Create a windows scheduler entry to run the script as often as required
' cscript -nologo PATH_WHERE_SCRIPT_STORED\jpgs_to_db.vbs

option explicit

dim TopFolder, MyDSN, MyTablename
dim FSO, ADO
 TopFolder="X:\photos"     ' <<---- change to suit your server
 MyDSN="mydbdsn"                   ' <<--- an ODBC DSN that points to your DB
 MyTablename="jpg_files"        ' <<--- change to suit your tablename
    ' create table  jpg_files(path varchar(50), name varchar(30), add_date datetime)
    ' then  in   sub processFile   change column-name references if necessary to suit

 Set FSO = CreateObject("Scripting.FileSystemObject")   
 Set ADO = CreateObject("ADODB.Connection")             
 ADO.ConnectionString= "DSN=" & MyDSN & ";"  

 ScanFolder topfolder
sub ScanFolder(path)
Dim folder, subFolders, fldr, subfiles, fil
  print "Fold: " & path
  Set folder = fso.GetFolder(path)

  set subfiles  = folder.files
  For Each fil in subfiles
    processFile path, fil

    ' --- process subfolders of current folder
  Set subFolders = folder.SubFolders
  For Each fldr in subFolders     
       call scanfolder(fldr)

End sub

sub processFile(path, fil)
Dim f, f1, fc, sql
  print "File: " & fil & "    Path: " & path & "    name: " &
  if right(lcase(fil),4)<>".jpg" then exit sub      '  a simple filter for JPGs only
                    ' --- change references to columns  path, name, add_date   to suit your table
  sql="select count(*) from " & MyTablename & " where path='" & path & "' and name='" & & "'"
  if getSql(sql)=0 then     ' add record
    sql = "insert into " & MyTablename & " (path, name, add_date) values('" & path & "', '" & & "', getdate() )"
  end if  
End sub

function getSql(sql)
dim recset
  Set recset = ADO.execute(sql)
  If not recset.EOF Then getSql=recset(0)
end function
sub print(s)
  wscript.echo s
end sub

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Very little is impossible,
what approx number of folders and files need to be catered for?
kenuk110Author Commented:
Thank you nrbreen!

Let me try this on my system, I really appreciate it, I'm not a programmer and haven't also used VB, the limited stuff I have done has been on C Sharp but let me change the parts you have suggested.

Really appreciate it.


kenuk110Author Commented:
Hi nrbreen,

Right, I have gone through the file, it all came in in one long string so I had to press enter a load of times, hence taking a little time to get back to you!

It all seems to be trying to do what it should but I have an issue where it requires the username and password for our SQL Server. Is there a way to add this information to the connection string as the message I get back is:

Login failed for user''.

Any ideas?
kenuk110Author Commented:
Hi again,

It works!!! I found the additional info on the net so added it and it works like a dream!

I really appreciate your help on this.

Best Regards,

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