This article is about to create a link to tables of another database. Microsoft Access has many built in function to do this task. But sometime we need to accomplish this task in programmatic way. Using this VBA function we are able to link with another database. In this article we are create an attached table in the current database from a table in a different MDB file.
To implement this function firstly we have to create the form with following controls as shown in Fig 1.1.
Fig:-1.1
On button click we have to implement the function. And after implementation we have to write the path of other database and the table of that database. As we press the linked button, table is link with current database as shown in fig 1.2.
Fig:-1.2
VBA code
Option Compare Database Private Sub Command0_Click() If IsNull(Me.txtdbname) Then MsgBox "Enter Data Base Path" Exit Sub End If If IsNull(Me.txttblname) Then MsgBox "Enter Table Name" Exit Sub End If If (createAttached(Me!txttblname, Me!txtdbname, Me!txttblname)) Then MsgBox ("Table is created") End If End Sub Function createAttached(strTable As String, strPath As String, strBaseTable As String) As Boolean On Error GoTo CreateAttachedError Dim tdf As TableDef, strConnect As String, fRetval As Boolean, myDB As Database DoCmd.SetWarnings False Set myDB = CurrentDb Set tdf = myDB.CreateTableDef(strTable) With tdf .Connect = ";DATABASE=" & strPath .SourceTableName = strBaseTable End With myDB.TableDefs.Append tdf fRetval = True DoCmd.SetWarnings True CreateAttachedExit: createAttached = fRetval Exit Function CreateAttachedError: If Err = 3110 Then Resume CreateAttachedExit Else If Err = 3011 Then Resume Next End If End If End Function
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