Access.2007.VBA.Bibl..

Create the table.

appAccess.DoCmd.RunSQL(strSQL)

ErrorHandlerExit:

Exit Sub

ErrorHandler:

AddInErr(Err)

Resume ErrorHandlerExit

End Sub

Public Sub CreateNewForm(ByVal control As _

Microsoft.Office.Core.IRibbonControl)

On Error GoTo ErrorHandler

Dim frm As Microsoft.Office.Interop.Access.Form

Dim txt As Microsoft.Office.Interop.Access.TextBox

Dim lbl As Microsoft.Office.Interop.Access.Label

Dim cbo As Microsoft.Office.Interop.Access.ComboBox

Dim lst As Microsoft.Office.Interop.Access.ListBox

Dim chk As Microsoft.Office.Interop.Access.CheckBox

Create a new form.

frm = appAccess.CreateForm()

frm.RecordSource = “tblTest”

txt = appAccess.CreateControl(FormName:=frm.Name, _

ControlType:=Microsoft.Office.Interop.Access.AcControlType.acTextBox, _

Section:=Microsoft.Office.Interop.Access.AcSection.acDetail, _

Left:=0, Top:=0, Width:=2500, Height:=400)

lbl = appAccess.CreateControl(FormName:=frm.Name, _

ControlType:=Microsoft.Office.Interop.Access.AcControlType.acLabel, _

Section:=Microsoft.Office.Interop.Access.AcSection.acDetail, _

Left:=0, Top:=1000, Width:=2500, Height:=400)

cbo = appAccess.CreateControl(FormName:=frm.Name, _

ControlType:=Microsoft.Office.Interop.Access.AcControlType.acComboBox, _

Section:=Microsoft.Office.Interop.Access.AcSection.acDetail, _

Left:=0, Top:=2000, Width:=2500, Height:=400)

lst = appAccess.CreateControl(FormName:=frm.Name, _

ControlType:=Microsoft.Office.Interop.Access.AcControlType.acListBox, _

Section:=Microsoft.Office.Interop.Access.AcSection.acDetail, _

Left:=0, Top:=3000, Width:=2500, Height:=400)

Customizing the Access Ribbon with a Visual Studio 2005 Shared Add-in 16