Can I make a group a member of a group?

(Walter Stabosz) #1

We are using Discourse as a project team communication tool. We’ve set it up as follows:

  • One Category for each project
  • One Group for each Category
  • We give project clients access to the Category, but we don’t want different clients to know about each other’s.
  • We have a Group staff2 to which all employees are added.
  • I called it staff2 because for some reason, I cannot add Users to to the default staff group, only admins, and they get added automatically.

I’d like for all of our employees to be able to see all Categories, but it’s cumbersome to have to manually add each user to each project Group. Is there a way I can add a Group as a member of another Group? i.e. add staff2 as a member of each project Group ?

I know that admins (and I think moderators) have access to all Categories, but I don’t want all of our employees to be admins.

(Jeff Atwood) #2

No, there is no concept of subgroups and we have no plans to add this at the moment.

(Bill Ayakatubby) #3

The “staff” group is for forum staff, not company/organization staff. Your “staff2” (“employees” might be better?) group is correct.

Can you not just give the “staff2” group access to each category directly? (Disclaimer: I don’t have a Discourse forum up and running, so I don’t technically know whether that’s possible. Just trying to be helpful.)

(Dave McClure) #4

Yep, this is the easiest way to handle what you’re trying to accomplish, @walterstabosz

You can ‘Add permission’ on a category to give additional groups access rights to it.

(Walter Stabosz) #5

@BhaelOchon & @mcwumbly : thanks for the suggestion to permission staff2 to each Category. This is exactly what I was looking for. @codinghorror: agreed, subgroups is an unnecessary feature.

(Kane York) #6

You should also name it something specific to your organization instead of staff2.

(Walter Stabosz) #7

Agreed. The group actually is called ssdel , not staff2. I just said staff2 so no one here would ask “what the hell is ssdel?”