At 01:15 2003-08-08 -0700, Bill Broadley wrote:
I have a mailserver(running postfix), I'd like it to .forward email
to my desktop (also running postfix).
So with a .forward, and I send a msg and I get:
"with a .forward". A .forward containing WHAT, exactly?
Perfect. Now on my desktop I can easily do something like:
:0
* ^From(_dot_)*random(_at_)foo(_dot_)com
proc/random
BTW, you're missing the LOCKFILE flag on the flags line. Also, that regexp
conditition will match the contents of the From: line as well as the From_
line, and the dot really should be escaped.
So that works great. Now if I want to replace .forward on the
server with a .procmailrc I tried (from the FAQ):
I take it that procmail is the LDA then, since it's perfectly legit to
invoke procmail (and thus, the .procmailrc) via a .forward...
:0c
! bill(_at_)desktop(_dot_)math(_dot_)ucdavis(_dot_)edu
The From_ of this message is going to be the account which forwarded the
message.
So now when I get a email:
.. via your desktop, I assume.
So I've lost the From header.
No, you still have a from header, it just doesn't reflect the original
message transmission, but rather the new message transmission. Look at a
listserve message sometime, and you'll see that the From_ is the list, not
the original sender.
This From_ header bit may be different depending upon the MTA you're using,
BTW.
You also have additional Received: lines, and potentially other headers
stuffed in by the receiving host, so it's hardly as if the message is an
absolute duplicate of the one first received at your upline host.
Am I missing something? I checked the archives, and various
FAQ's without finding a way to preserve the From header just
like .forward does.
:0
* ^From \/[^ ]*
| $SENDMAIL my-other-address -f${MATCH}
The exact commandline invocation may differ depending upon the MTA you're
running. The above is for sendmail. Well behaved alternatives generally
provide a sendmail lookalike submission program though.
Your MTA may end up stuffing an X-Authentication-Warning into the headers
of the forwarded message if you are an untrusted user.
Ideas?
x^n + y^n = z^n has no whole number solution when n is greater than
2. Not my idea by any means, but it's an idea which you might be able to
appreciate.
---
Sean B. Straw / Professional Software Engineering
Procmail disclaimer: <http://www.professional.org/procmail/disclaimer.html>
Please DO NOT carbon me on list replies. I'll get my copy from the list.
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