[Mailman-Users] List all members of All lists

[Adam McGreggor]

On Fri, Nov 20, 2009 at 02:10:06PM -0500, Harold Pritchett wrote:
> How do I list all of the members of all of my lists.
> 
> Actually, I need to delete all users from a domain
> and the command "remove_members --fromall --nouserack"
> does not accept wildcards.
> 
> If I can get a list of all users, I can then grep it
> for the domain name and run remove_members for each
> of the users which are found.

I'd probably use something like 
    find_member example.com > outfile
which will give you the email address and list memberships; this can
be fed to remove_members. That may need some tidy-up, though.

Another approach I might consider would be to use list_lists -b, and a
while loop; something like: (untested)

        list_lists -b | while read L
        do
            find_member -l ${L} example.com > found-${L}
            remove_members --fromall --nouserack -f found-${L} ${L}
            rm found-${L}
        done

Maybe skipping the outfile, but I like outfiles, rather than
pipes/redirects.

maybe feeding stuff through tee(1), although logs/subscribe should log
those, appropriately.

(I'd probably use the second suggestion, myself, having tested it
first with a regexp that wouldn't do much damage.)

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    A place where minutes are kept and hours are lost.''