MS Access application return value - ms-access

Question

I need to make automatic application called from windows command line. This app start, make some DB operations and automatically exit.
Next I need to handle return values from Access. I tried everything what I find on Google but Access returns 0 every time. I need to return 1 or something else to tell cmd that there is error in database.
here is some code in VBA (access call Form when application starts):
Private Sub Form_Load()
Dim xSQL As String
Dim xRS As DAO.Recordset
xSQL = "select * from table1"
Set xRS = CurrentDb.OpenRecordset(xSQL, dbOpenDynaset)
xRS.MoveLast: xRS.MoveFirst
DoEvents
If xRS!field1 = "" Or IsNull(xRS!field1) Then
' HERE I WANT TO RETURN 1 OR ANY TRUE VALUE = ERROR
Application.Quit acQuitSaveAll
End If
' HERE I RETURN 0 = SUCCESS
Application.Quit acQuitSaveAll
End Sub
Some ideas how to do it?

Answer 1

I think your best solution can be setting an environment variable with the desired return value.
From your VBA code:
Private Declare Function SetEnvironmentVariable Lib "kernel32" _
Alias "SetEnvironmentVariableA" _
(ByVal lpName As String, _
ByVal lpValue As String) As Long
Private Sub Form_Load()
Dim xSQL As String
Dim xRS As DAO.Recordset
xSQL = "select * from table1"
Set xRS = CurrentDb.OpenRecordset(xSQL, dbOpenDynaset)
xRS.MoveLast: xRS.MoveFirst
DoEvents
If xRS!field1 = "" Or IsNull(xRS!field1) Then
SetEnvironmentVariable "ACCESSEXITCODE", "1"
Application.Quit acQuitSaveAll
End If
SetEnvironmentVariable "ACCESSEXITCODE", "0"
Application.Quit acQuitSaveAll
End Sub
From de command line:
IF %ACCESSEXITCODE%==1 GOTO HandleError