[Mailman-Users] Re: Question on outputting all lists and list owners and a process for batch disabling of lists without an owner

On 5/27/22 10:22, John Lake wrote:

Our security department has asked me to canvas all of our lists (4,237!)—and if a list owner value is not defined they want me to disable the list.

1. Can I combine the list_lists and list_owners commands to output both (hopefully linked) values as a .csv?

list_owners -w will list the owners by list name. If a list has no owners, only the list name will be output for that list. You could then process that file to identify the lists with no owners.

2. Is there a script that any of you have leveraged to batch disable lists without a defined owner?

What do you mean by disable?

A withlist script like

def no_owner(mlist)
     if len mlist.owner == 0:
         if not mlist.Locked:
             mlist.Lock
         print('List %s has no owners, disabling.' % mlist.real_name)
         <code here to disable list. I could provide it if I knew what 
that meant>
         mlist.Save()
         mlist.Unlock()

saved as no_owner.py in Mailman's bin/ directory and run via

bin/withlist -a -r no_owner

You could omit the code

         if not mlist.Locked:
             mlist.Lock
         <code here to disable list. I could provide it if I knew what 
that meant>
         mlist.Save()
         mlist.Unlock()

I.e., just leave the print statement in the if len mlist.owner == 0: clause to just print the names of lists with no owner.

-- Mark Sapiro <mark@msapiro.net> The highway is for gamblers, San Francisco Bay Area, California better use your sense - B. Dylan