python imaplib获取gmail收件箱主题标题和发件人姓名

Question

我正在使用 pythons imaplib 连接到我的 Gmail 帐户。我想检索前 15 条邮件（未读或已读，没关系）并仅显示主题和发件人姓名（或地址），但不知道如何显示收件箱的内容。

这是到目前为止我的代码（成功连接）

import imaplib

mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login('[email protected]', 'somecrazypassword')
mail.list()
mail.select('inbox')

#need to add some stuff in here

mail.logout()

我相信这应该足够简单，我只是对 imaplib 库的命令不够熟悉。任何帮助将不胜感激...

更新 感谢 Julian，我可以遍历每条消息并使用以下命令检索整个内容：

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()

但我只想要主题和发件人。是否有针对这些项目的 imaplib 命令，或者我必须解析 data[0][1] 的全部内容以获取文本：主题和发件人？

更新 好的，主题和发件人部分正常工作，但迭代 (1, 15) 是按 desc 顺序完成的，显然首先向我显示最旧的消息。我怎样才能改变这个？我尝试这样做：

for i in range( len(data[0])-15, len(data[0]) ):
     print data

但这只是给了我所有 15 次迭代的 None...有什么想法吗？我也尝试过 mail.sort('REVERSE DATE', 'UTF-8', 'ALL') 但 gmail 不支持 .sort() 函数

UPDATE 找到了一种方法：

#....^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

这给了我最新的 15 条消息主题和发件人地址，按要求的降序排列！感谢所有提供帮助的人！

Answer 1

import email

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].decode())
            for header in [ 'subject', 'to', 'from' ]:
                print('%-8s: %s' % (header.upper(), msg[header]))

这应该会让您了解如何检索主题以及从哪里检索主题？

Answer 2

这是我从电子邮件中获取有用信息的解决方案：

import datetime
import email
import imaplib
import mailbox


EMAIL_ACCOUNT = "[email protected]"
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

Answer 3

对于那些寻找如何检查邮件和解析标头的人，这就是我使用的：

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

Answer 4

我正在寻找一个现成的简单脚本来通过 IMAP 列出最后一个收件箱，而无需对所有邮件进行排序。这里的信息很有用，尽管 DIY 并遗漏了一些方面。首先， IMAP4.select 返回消息计数。其次，主题标头解码并不简单。

#! /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('[email protected]', '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)

Answer 5

BODY 获取几乎所有内容并将消息标记为已读。
BODY[] 仅获取这些部分。
BODY.PEEK[] 获取相同的部件，但不标记消息已读。
可以是 HEADER 或 TEXT 或 HEADER.FIELDS（<字段列表>） 或
HEADER.FIELDS.NOT (<字段列表>)

这就是我使用的： typ, data = connection.fetch(message_num_s, b'(BODY.PEEK[HEADER.FIELDS ( SUBJECT FROM)])')

`

def safe_encode(seq):
    if seq not in (list,tuple):
        seq = [seq]
    for i in seq:
        if isinstance(i, (int,float)):
            yield str(i).encode()
        elif isinstance(i, str):
            yield i.encode()
        elif isinstance(i, bytes):
            yield i
        else:
            raise ValueError

def fetch_fields(connection, message_num, field_s):
    """Fetch just the fields we care about. Parse them into a dict"""
    if isinstance(field_s, (list,tuple)):
        field_s = b' '.join(safe_encode(field_s))
    else:
        field_s = tuple(safe_encode(field_s))[0]

    message_num = tuple(safe_encode(message_num))[0]

    typ, data = connection.fetch(message_num, b'(BODY.PEEK[HEADER.FIELDS (%s)])'%(field_s.upper()))
    if typ != 'OK':
        return typ, data  #change this to an exception if you'd rather

    items={}
    lastkey = None
    for line in data[0][1].splitlines():
        if b':' in line:
            lastkey, value = line.strip().split(b':', 1)
            lastkey = lastkey.capitalize()
            #not all servers capitalize the same, and some just leave it
            #as however it arrived from some other mail server.

            items[lastkey]=value
        else:
            #subject was so long it ran onto the next line, luckily it didn't have a ':' in it so its easy to recognize.
            items[lastkey]+=line
            #print(items[lastkey])
    return typ, items
`

您将其放入代码示例中：通过将对 'mail.fetch()' 的调用替换为 fetch_fields(mail, i, 'SUBJECT FROM') 或 fetch_fields(mail, i, ('SUBJECT' 'FROM'))

Answer 6

添加到上述所有答案。

import imaplib
import base64
import os
import email

if __name__ == '__main__':
    email_user = "[email protected]"
    email_pass = "********"
    mail = imaplib.IMAP4_SSL("hostname", 993)
    mail.login(email_user, email_pass)
    mail.select()
    type, data = mail.search(None, 'ALL')
    mail_ids = data[0].decode('utf-8')
    id_list = mail_ids.split()
    mail.select('INBOX', readonly=True)
    for i in id_list:
        typ, msg_data = mail.fetch(str(i), '(RFC822)')
        for response_part in msg_data:
            if isinstance(response_part, tuple):
                msg = email.message_from_bytes(response_part[1])
                print(msg['from']+"\t"+msg['subject'])

这将为您提供电子邮件的发件人和主题名称。