Re: [Mailman-Users] Removing invalid list members

On Fri, Dec 01, 2006 at 11:03:26AM -0800, Mark Sapiro wrote: > Barry Warsaw wrote: > >What I tried to say was, use "bin/withlist -l mylist" and call > >m.removeMember() explicitly. Then m.Save() to save the changes. Yep, that took care of it. Thanks! > And, please note that this is in the FAQ

Re: [Mailman-Users] Removing invalid list members

Barry Warsaw wrote: > >What I tried to say was, use "bin/withlist -l mylist" and call >m.removeMember() explicitly. Then m.Save() to save the changes. And, please note that this is in the FAQ at . -- Mark Sapiro <[EMAIL PRO

Re: [Mailman-Users] Removing invalid list members

-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 [Barry swears he typed a response!] What I tried to say was, use "bin/withlist -l mylist" and call m.removeMember() explicitly. Then m.Save() to save the changes. - -Barry -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.5 (Darwin) iQCVAwUBRXBr

Re: [Mailman-Users] Removing invalid list members

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Re: [Mailman-Users] Removing invalid list members

Three days and no responses... So there's no way to remove the address described below? The code that strips the second @ and everything after it can't be bypassed? You can't safely edit the database to delete it directly without going through remove_members and the sanitizing code? There's got

[Mailman-Users] Removing invalid list members

Somehow, a user has gotten himself subscribed to one of my lists with an address of the form [EMAIL PROTECTED]@hotmail.com. How do I get rid of this subscription? The various web interface options just give an error that this address is not subscribed (even while showing it on the list of subscri