Option Box commands

[AT76]

Hello, I have created an option box (Option87)in my access form and would like to make it work as follows:

If Option87 is pressed Then Do.....

Could someone help me with the syntax. I tried using:

(If Option87.Onclick ...) but it did not work.

Thank you!

 

[comaboy]

You need to understand the Event model - it looks like you're about to program an endless loop checking for a change in the interface. In fact, it's all done for you. Bring up the Form in Design View and then double-click your Option Control. You'll then jump to the Code Module, and Access will automatically build an "OnClick" function for you. (probably something like..)

Sub Option87_OnClick()

End Sub

You can then place your code in the event procedure.

Hope this helps!

 

[ZmrAbdulla]

I have created an option box (Option87)in my access form and would like to make it work as follows:

Is this Access or .NET?
If it is Access then here is something

Private Sub MychkBox_AfterUpdate()
    Select Case Me.MychkBox.Value
    Case -1
        Me.MyTextBox = Now
    Case 0
        Me.MyTextBox = ""
    End Select
End Sub

'=====================================
Private Sub MychkBox_AfterUpdate()
    If Me.MychkBox.Value = True Then
        Me.MyTextBox = Now
    Else
        Me.MyTextBox = ""
    End If
End Sub

You can do it either way.

________________________________________________________
Zameer Abdulla
Help to find Missing people
Do not cut down the tree that gives you shade.

 

[AT76]

Thank you ZmrAbdulla! I will try what you suggest!

Best regards,

AT