Get message header (subject) with poplib python

Question

I want get only the subject from a message mail using poplib:

import poplib
server ='pop3.live.com'
port = 995
login="xxx@outlook.com"
pw="xxx"
print "Connecting..."
M = poplib.POP3_SSL(server,port)
print "Connected to "+server
M.user(login)
M.pass_(pw)
print login+" is authenticated."
numMessages = len(M.list()[1])
for i in range(numMessages):
    for j in M.retr(i+1)[1]:
        print j
M.quit()

Any suggestions? Thanks in advance

This question/answer seems to answer your question [http://stackoverflow.com/questions/4908472/how-to-recieve-mail-using-python][1] [1]: http://stackoverflow.com/questions/4908472/how-to-recieve-mail-using-python — DisappointedByUnaccountableMod, May 22 '13 at 10:49

score 6 · Answer 1 · answered Oct 06 '16 at 14:55

You need to use email.message_from_bytes to convert e-mail from bytes and decode_header to decode your headers. This is python 3.4 example. In python 2 code will be different.

#!/usr/bin/env python3

import email, poplib
from email.header import decode_header

login = 'login'
password = 'password'
pop_server = 'gmail.com'
pop_port = 995

mail_box = poplib.POP3_SSL(pop_server, pop_port)
mail_box.user(login)
mail_box.pass_(password)

num_messages = len(mail_box.list()[1])
for i in range(num_messages):
    print(i, "message:")

    raw_email  = b"\n".join(mail_box.retr(i+1)[1])
    parsed_email = email.message_from_bytes(raw_email)

    subject = decode_header(parsed_email['Subject'])
    print(
        subject[0][0].decode(subject[0][1])
    )

mail_box.quit()

Get message header (subject) with poplib python

1 Answers1