Re: FW: Re: Looking for package to manipulate pop3 mailbox index
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
On Tuesday 08 January 2002 10:53 pm, dman wrote:
> Ron : use the "reply to all" button (or whatever kmail has) to keep
> the replies on-list.
Sorry. The other lists I belong to munge Reply-to, so it's a
pretty ingrained habit to "reply to sender". I do it right approx
1/3 of the time...
> ----- Forwarded message from Ron Johnson <ron.l.johnson@home.com>
> -----
>
> From: Ron Johnson <ron.l.johnson@home.com>
> To: dman <dsh8290@ritvax.rit.edu>
> Subject: Re: Looking for package to manipulate pop3 mailbox index
> Date: Tue, 08 Jan 2002 21:40:09 -0600
> X-Mailer: KMail [version 1.3.1]
>
> On Tuesday 08 January 2002 05:34 pm, dman wrote:
> > On Tue, Jan 08, 2002 at 04:59:15PM -0600, Ron Johnson wrote:
> > | On Tuesday 08 January 2002 04:38 pm, Dougie Nisbet wrote:
> > | > On another ISP I've got a full mailbox (around 120MB and 15000
> > | > messages I think) because a spammer has faked a legitimate
> > | > e-mail address of mine and used it. This means that
> > | > 'lunsform3459@nisbet.compulink.co.uk' (which I had no idea was
> > | > legitimate) has received many thousands of mail bounces.
> > | >
> > | > Is there a package that would allow me to examine JUST THE
> > | > HEADERS of these mail messages, and allow me to delete chunks
> > | > at a time? There may be a handful of mails in there that I
> > | > want, but I don't want to download 120MB to find out!
> > |
> > | I'm 99.9% sure you can write something in Python.
> >
> > Of course you can! :-). (or just use telnet after reading the
> > POP3 RFC)
>
> And here's the python script:
> import getpass, poplib, string
> M = poplib.POP3('mail.foobar.com')
> M.user('your.address')
> M.pass_(getpass.getpass())
> numMessages = len(M.list()[1])
> print numMessages
> x = M.list(0)
> for i in range(1, numMessages):
> print '=============================================='
> print '=============================================='
> x = M.top(i, 0)
> print i
> for y in x[1]:
> z = string.split(y, ' ')
> if z[0] in ('Date:', 'From:', 'To:', 'Subject:'):
> print z
> a = raw_input('Delete this message?')
> if a in ('Y', 'y'):
> M.dele(i)
> if a in ('Q', 'q'):
> break
> M.quit()
- --
+------------------------------------------------------------+
| Ron Johnson, Jr. Home: ron.l.johnson@home.com |
| Jefferson, LA USA http://ronandheather.dhs.org:81 |
| |
! "Fair is where you take your cows to be judged." !
! Unknown !
+------------------------------------------------------------+
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.0.6 (GNU/Linux)
Comment: For info see http://www.gnupg.org
iD4DBQE8O88gjTz5dS9Us5wRAufRAJjvSdp+eBZF/hOetSILrHkE/z0SAJ9vacAO
8jxq4sN/DYHc8imnHxd4Og==
=Q3Vl
-----END PGP SIGNATURE-----
Reply to: