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# Copyright (C) 1998,1999,2000,2001,2002 by the Free Software Foundation, Inc.
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# Copyright (C) 1998-2005 by the Free Software Foundation, Inc.
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# This program is free software; you can redistribute it and/or
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# modify it under the terms of the GNU General Public License
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# as published by the Free Software Foundation; either version 2
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# of the License, or (at your option) any later version.
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# This program is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU General Public License for more details.
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# You should have received a copy of the GNU General Public License
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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA.
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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301,
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"""Determine whether the message is approved for delivery.
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determines whether the message is definitively approved or definitively
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denied. Situations that could hold a message for approval or confirmation are
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not tested by this module.
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from email.Iterators import typed_subpart_iterator
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from Mailman import mm_cfg
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from Mailman import Errors
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if passwd is missing:
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# Find the first text/plain part in the message
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for part in typed_subpart_iterator(msg, 'text', 'plain'):
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# XXX I'm not entirely sure why, but it is possible for the payload of
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# the part to be None, and you can't splitlines() on None.
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if part is not None and part.get_payload() is not None:
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lines = part.get_payload().splitlines()
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for lineno, line in zip(range(len(lines)), lines):
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# Now strip the first line from the payload so the
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# password doesn't leak.
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part.set_payload(NL.join(lines[1:]))
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part.set_payload(NL.join(lines))
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# MAS: Bug 1181161 - Now try all the text parts in case it's
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# multipart/alternative with the approved line in HTML or other
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# text part. We make a pattern from the Approved line and delete
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# it from all text/* parts in which we find it. It would be
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# better to just iterate forward, but email compatability for pre
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# Python 2.2 returns a list, not a true iterator.
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# This will process all the multipart/alternative parts in the
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# message as well as all other text parts. We shouldn't find the
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# pattern outside the mp/a parts, but if we do, it is probably
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# best to delete it anyway as it does contain the password.
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# Make a pattern to delete. We can't just delete a line because
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# line of HTML or other fancy text may include additional message
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# text. This pattern works with HTML. It may not work with rtf
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# or whatever else is possible.
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pattern = name + ':(\s| )*' + re.escape(passwd)
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for part in typed_subpart_iterator(msg, 'text'):
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if part is not None and part.get_payload() is not None:
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# Should we decode the payload?
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lines = part.get_payload()
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if re.search(pattern, lines):
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part.set_payload(re.sub(pattern, '', lines))
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if passwd is not missing and mlist.Authenticate((mm_cfg.AuthListModerator,
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mm_cfg.AuthListAdmin),
added
removed
Mailman/Handlers/Approve.py
