parsing through log file and printing out regex backreferences

Question

I want to parse my /var/log/maillog to print out the email addresses in "from=<>" and "to=<>" (mainly to do a quick check for false positives in the DNS RBLs that I've configured)

A maillog entry looks like this (postfix):

Jun 20 17:27:52 foobarserver postfix/smtpd[15925]: NOQUEUE: reject: RCPT from sbr.nouveauquebec.com[184.22.154.110]: 554 5.7.1 Service unavailable; Sender address [corporate@nouveauquebec.com] blocked using urired.spameatingmonkey.net; Red listed, see http://spameatingmonkey.com/lookup/nouveauquebec.com; from=<something@spam.com> to=<foo@foo.bar> proto=ESMTP helo=<sbr.nouveauquebec.com>

and my regex is

from=<(\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b)> to=<(\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b)>

I'm storing the matches in back references \1 and \2 which I want to print out. What I want to do is pipe the output from cat /var/log/maillog and apply the regex on each line and output the backreferences.

Is there a quick way to accomplish this?

Answer 1

Using sed:

sed -r -n 's/^.* from=<([^>]+)>\s*to=<([^>]+)>.*$/\1 \2/p' /var/log/maillog

Answer 2

This one uses a simplified regex and non-greedy quantifiers:

perl -ne 's/^.*from=<(.+?@.+?)> to=<(.+?@.+?)>.*$/$1 $2/; print;' /var/log/maillog