Roy,
By using the BitOr function, you can basically include as many groups as
you have. It "should" filter out any duplicate permission you may have.
Just off the cuff,
Loop through your groups and do a BitOr comparison of the first two
groups. Take the result and do a BitOr on the next group and so on. The
final result should be the composite permission for that individual.
Good Luck!
BTW, Let's take this conversation to the fbcommunity list!
Bob Krieger
www.FBTutor.com
The FuseBox Tutor
----- Original Message -----
From: "R Vosmeer" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 10, 2002 8:03 AM
Subject: RE: Another cf_secure question
> Thanks Bob,
>
> I am using multiple permission files so I will place them in the
> circuits and include them there.
>
> I read the document and I see that it is possible to use the BitOr when
> a user belongs to two usergroups. But is it possible to let the user
> belong to more than two groups??
>
> -----Oorspronkelijk bericht-----
> Van: Bob Krieger [mailto:[EMAIL PROTECTED]]
> Verzonden: 10 June 2002 15:44
> Aan: [EMAIL PROTECTED]
> Onderwerp: Re: Another cf_secure question
>
> Roy,
> If you have all your permissions defined in ONE fbx_Permissions
> file,
> you should only need to include it within the ROOT fbx_settings file. If
> each fbx_Permissions file defines ONLY those premissions used within
> each
> specific circuit, then each one should be included ONLY in that
> particular
> circuit settings file.
>
> Bob Krieger
> www.FBTutor.com
> The FuseBox Tutor
>
>
> ----- Original Message -----
> From: "R Vosmeer" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Monday, June 10, 2002 7:33 AM
> Subject: Another cf_secure question
>
>
> > Hi,
> >
> > I am setting up a system in which i want to implement the Hal Helms
> > cf_secure tag. I am using the BIT model and the fbx_permissions.cfm
> > file. Where do you people cfinclude this file? I now have it set up in
> > the settings file of each circuit. Is this correct or is there a
> better
> > way of doing this.
> >
> > Thanks..
> >
> > Roy
> >
> >
> >
>
>
>
>
>
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