问题
I'm using pythons imaplib to connect to my gmail account. I want to retrieve the top 15 messages (unread or read, it doesn't matter) and display just the subjects and sender name (or address) but don't know how to display the contents of the inbox.
Here is my code so far (successful connection)
import imaplib
mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login('mygmail@gmail.com', 'somecrazypassword')
mail.list()
mail.select('inbox')
#need to add some stuff in here
mail.logout()
I believe this should be simple enough, I'm just not familiar enough with the commands for the imaplib library. Any help would be must appreciated...
UPDATE thanks to Julian I can iterate through each message and retrieve the entire contents with:
typ, data = mail.search(None, 'ALL')
for num in data[0].split():
typ, data = mail.fetch(num, '(RFC822)')
print 'Message %s\n%s\n' % (num, data[0][1])
mail.close()
but I'm wanting just the subject and the sender. Is there a imaplib command for these items or will I have to parse the entire contents of data[0][1] for the text: Subject, and Sender?
UPDATE OK, got the subject and sender part working but the iteration (1, 15) is done by desc order apparently showing me the oldest messages first. How can I change this? I tried doing this:
for i in range( len(data[0])-15, len(data[0]) ):
print data
but that just gives me None
for all 15 iterations... any ideas? I've also tried mail.sort('REVERSE DATE', 'UTF-8', 'ALL')
but gmail doesnt support the .sort() function
UPDATE Figured out a way to do it:
#....^other code is the same as above except need to import email module
mail.select('inbox')
typ, data = mail.search(None, 'ALL')
ids = data[0]
id_list = ids.split()
#get the most recent email id
latest_email_id = int( id_list[-1] )
#iterate through 15 messages in decending order starting with latest_email_id
#the '-1' dictates reverse looping order
for i in range( latest_email_id, latest_email_id-15, -1 ):
typ, data = mail.fetch( i, '(RFC822)' )
for response_part in data:
if isinstance(response_part, tuple):
msg = email.message_from_string(response_part[1])
varSubject = msg['subject']
varFrom = msg['from']
#remove the brackets around the sender email address
varFrom = varFrom.replace('<', '')
varFrom = varFrom.replace('>', '')
#add ellipsis (...) if subject length is greater than 35 characters
if len( varSubject ) > 35:
varSubject = varSubject[0:32] + '...'
print '[' + varFrom.split()[-1] + '] ' + varSubject
this gives me the most recent 15 message subject and sender address in decending order as requested! Thanks to all who helped!
回答1:
c.select('INBOX', readonly=True)
for i in range(1, 30):
typ, msg_data = c.fetch(str(i), '(RFC822)')
for response_part in msg_data:
if isinstance(response_part, tuple):
msg = email.message_from_string(response_part[1])
for header in [ 'subject', 'to', 'from' ]:
print '%-8s: %s' % (header.upper(), msg[header])
This should give you an idea on how to retrieve the subject and from?
回答2:
For those looking for how to check mail and parse the headers, this is what I used:
def parse_header(str_after, checkli_name, mailbox) :
#typ, data = m.search(None,'SENTON', str_after)
print mailbox
m.SELECT(mailbox)
date = (datetime.date.today() - datetime.timedelta(1)).strftime("%d-%b-%Y")
#date = (datetime.date.today().strftime("%d-%b-%Y"))
#date = "23-Jul-2012"
print date
result, data = m.uid('search', None, '(SENTON %s)' % date)
print data
doneli = []
for latest_email_uid in data[0].split():
print latest_email_uid
result, data = m.uid('fetch', latest_email_uid, '(RFC822)')
raw_email = data[0][1]
import email
email_message = email.message_from_string(raw_email)
print email_message['To']
print email_message['Subject']
print email.utils.parseaddr(email_message['From'])
print email_message.items() # print all headers
回答3:
This was my solution to get the useful bits of information from emails:
import datetime
import email
import imaplib
import mailbox
EMAIL_ACCOUNT = "your@gmail.com"
PASSWORD = "your password"
mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login(EMAIL_ACCOUNT, PASSWORD)
mail.list()
mail.select('inbox')
result, data = mail.uid('search', None, "UNSEEN") # (ALL/UNSEEN)
i = len(data[0].split())
for x in range(i):
latest_email_uid = data[0].split()[x]
result, email_data = mail.uid('fetch', latest_email_uid, '(RFC822)')
# result, email_data = conn.store(num,'-FLAGS','\\Seen')
# this might work to set flag to seen, if it doesn't already
raw_email = email_data[0][1]
raw_email_string = raw_email.decode('utf-8')
email_message = email.message_from_string(raw_email_string)
# Header Details
date_tuple = email.utils.parsedate_tz(email_message['Date'])
if date_tuple:
local_date = datetime.datetime.fromtimestamp(email.utils.mktime_tz(date_tuple))
local_message_date = "%s" %(str(local_date.strftime("%a, %d %b %Y %H:%M:%S")))
email_from = str(email.header.make_header(email.header.decode_header(email_message['From'])))
email_to = str(email.header.make_header(email.header.decode_header(email_message['To'])))
subject = str(email.header.make_header(email.header.decode_header(email_message['Subject'])))
# Body details
for part in email_message.walk():
if part.get_content_type() == "text/plain":
body = part.get_payload(decode=True)
file_name = "email_" + str(x) + ".txt"
output_file = open(file_name, 'w')
output_file.write("From: %s\nTo: %s\nDate: %s\nSubject: %s\n\nBody: \n\n%s" %(email_from, email_to,local_message_date, subject, body.decode('utf-8')))
output_file.close()
else:
continue
回答4:
I was looking for a ready made simple script to list last inbox via IMAP without sorting through all messages. The information here is useful, though DIY and misses some aspects. First, IMAP4.select returns message count. Second, subject header decoding isn't straightforward.
#! /usr/bin/env python
# -*- coding: utf-8 -*-
import imaplib
import email
from email.header import decode_header
import HTMLParser
# to unescape xml entities
_parser = HTMLParser.HTMLParser()
def decodeHeader(value):
if value.startswith('"=?'):
value = value.replace('"', '')
value, encoding = decode_header(value)[0]
if encoding:
value = value.decode(encoding)
return _parser.unescape(value)
def listLastInbox(top = 4):
mailbox = imaplib.IMAP4_SSL('imap.gmail.com')
mailbox.login('mygmail@gmail.com', 'somecrazypassword')
selected = mailbox.select('INBOX')
assert selected[0] == 'OK'
messageCount = int(selected[1][0])
for i in range(messageCount, messageCount - top, -1):
reponse = mailbox.fetch(str(i), '(RFC822)')[1]
for part in reponse:
if isinstance(part, tuple):
message = email.message_from_string(part[1])
yield {h: decodeHeader(message[h]) for h in ('subject', 'from', 'date')}
mailbox.logout()
if __name__ == '__main__':
for message in listLastInbox():
print '-' * 40
for h, v in message.items():
print u'{0:8s}: {1}'.format(h.upper(), v)
来源:https://stackoverflow.com/questions/7314942/python-imaplib-to-get-gmail-inbox-subjects-titles-and-sender-name