Here is a solution I received from Jerry Stratton using withlist command
from the command line...
Create a file to use with "withlist".
==============snip=========================
from Mailman import mm_cfg
def unmoderated(mlist):
memberCount = 0
for member in mlist.getMembers():
if not mlist.getMemberOption(member, mm_cfg.Moderate):
print member
memberCount = memberCount + 1
print "Unmoderated members found:", memberCount
==============snip=========================
If this is called "queries.py", then you can use the following command to
list all unmoderated members of a mailing list:
withlist -r queries.unmoderated LISTNAME
Note that the file queries.py must be located in the home directory
of mailman (not in bin).
This is exactly what I needed. Thanks to Jerry for this solution.
--Donald
------------------------------------------------------
Mailman-Users mailing list
Mailman-Users@python.org
http://mail.python.org/mailman/listinfo/mailman-users
Mailman FAQ: http://wiki.list.org/x/AgA3
Searchable Archives: http://www.mail-archive.com/mailman-users%40python.org/
Unsubscribe:
http://mail.python.org/mailman/options/mailman-users/archive%40jab.org
Security Policy: http://wiki.list.org/x/QIA9
