This MSAccess tutorial explains how to create a sequential number that you control instead of using an autonumber field in Access 2003 (with screenshots and step-by-step instructions).
Question: In Microsoft Access 2003/XP/2000/97, I want to create a sequential number that I can control instead of using an AutoNumber field. How can I do this?
Answer: We've created a sample Access database that you can download that demonstrates how to create a sequential number that you can control.
Let's look at the example. Below, we have a form that allows the user to enter a record. The user will enter the Date, Description, and Value, and the form will automatically assign a sequential number to the Journal # field.
The user can control the assignment of the next number through the Codes table. In this table, there is a record for Journal Number that displays the last number assigned. The user can modify this value accordingly to start at whatever number is appropriate.
Then in Module1 in the Access database, there is a function called NewJournalNbr that returns the next number in the sequence and increments the Last_Nbr_Assigned field by 1.
Function NewJournalNbr() As Long Dim db As Database Dim LSQL As String Dim LUpdate As String Dim Lrs As DAO.Recordset Dim LNewJournalNbr As Long On Error GoTo Err_Execute Set db = CurrentDb() 'Retrieve last number assigned for Journal Number LSQL = "Select Last_Nbr_Assigned from Codes" LSQL = LSQL & " where Code_Desc = 'Journal Number'" Set Lrs = db.OpenRecordset(LSQL) 'If no records were found, return an error If Lrs.EOF = True Then LNewJournalNbr = 0 MsgBox "There was no entry found in the Codes table for Journal Number." Else 'Determine new Journal Number LNewJournalNbr = Lrs("Last_Nbr_Assigned") + 1 'Increment Journal Number in Codes table by 1 LUpdate = "Update Codes" LUpdate = LUpdate & " set Last_Nbr_Assigned = " & LNewJournalNbr LUpdate = LUpdate & " where Code_Desc = 'Journal Number'" db.Execute LUpdate, dbFailOnError End If Lrs.Close Set Lrs = Nothing Set db = Nothing NewJournalNbr = LNewJournalNbr Exit Function Err_Execute: 'An error occurred, return 0 NewJournalNbr = 0 MsgBox "An error occurred while trying to determine the next Journal Number to assign." End Function
If after trying this example, you receive a "not defined" error on the "Dim db as Database" declaration, you will need to follow some additional instructions.
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