python - Print only specific part of email -

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i have code (below) shows me emails in email account. shows whole email, including metadata (which dont want). there way print to, from, subject , message only? in python well. thanks.

code:

import getpass, imaplib import os  email = raw_input('email: ') password = getpass.getpass() m = imaplib.imap4_ssl("imap.gmail.com", 993) print('logging in ' + email + '...') m.login(email, password) m.select() typ, data = m.search(none, 'all') num in data[0].split():     typ, data = m.fetch(num, '(rfc822)')     print ('message %s\n%s\n' % (num, data[0][1])) m.close() m.logout()

you can use email.parser.parser() standard module parse mail , headers 

from __future__ import print_function import imaplib import getpass import os email.parser import parser  email = raw_input('email: ') password = getpass.getpass()  print('logging in as', email, '...')  m = imaplib.imap4_ssl("imap.gmail.com", 993) m.login(email, password)  m.select() typ, data = m.search(none, 'all')  num in data[0].split():     typ, data = m.fetch(num, '(rfc822)')     #print ('message %s\n%s\n' % (num, data[0][1]))      header = parser().parsestr(data[0][1])     print('from:', header['from'])     print('to:', header['to'])     print('subject:', header['subject'])      print('body:')     part in header.get_payload():         print(part.as_string()[:150], '.....')      #break # test first message  m.close() m.logout()

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