>>First, thank you for your suggestions. I get it that I need to create a container control with three option buttons. This is fine.
>>
>>But I don't understand why there are 8 possibilities. How do you get the value of 7?
>
>7 is when all options are set (this is your case 6 in the list)
>
>You forgot to add Shift3 (alone) to your list
>
>each shift is a bit that can be on or off. They are bits 2, 1 and 0 of an integer
>
>__
You are absolutely correct. I missed that. Thank you for your help.
>
>>>(1) I think there are 8 possibilities - from 0 to 7
>>>shift1 = 2 ^0
>>>shift2 = 2^1
>>>shift3 = 2^ 2
>>>
>>>
>>>(2) I'm afraid you'll have to use a container with 3 option buttons
>>>
>>>(3) Add an unvisible spinner to the container (range 0-7) and set the controlSource of the spinner to the field
>>>
>>>(4) add code to the container
>>> - refresh = setting the options on/off (easy with bittest(field, 0) , bittest(field, 1) , bittest(field, 2)
>>>
>>>(5) add code to the optionbuttons (interactive change) to set the value of the spinner accordingly
>>>
>>>_
>>>
>>>>Hi,
>>>>
>>>>I need to create a control that will have 3 choices kind of similar to how the option group control works but use should be able to select more than 1 choice. For example:
>>>>
>>>>Shift 1 [ ] Shift 2 [ ] Shift 3 [ ]
>>>>
>>>>The control should be bound to one field, of Int type. So that the value stored in the field would be as following:
>>>>0 - if user didn't make any selection
>>>>1 - If user selected Shift 1
>>>>2 - Shift2
>>>>3 - Shift1 + Shift2
>>>>4 - Shift1 + Shift3
>>>>5 - Shift2 + Shift3
>>>>6 - Shift1 + Shift2 + Shift3
>>>>
>>>>The only way I can think of is by creating a container-based class with three optio control buttons.
>>>>
>>>>But, in case I am missing something, is there a way to use one native VFP control for this?
>>>>
>>>>TIA.
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