如何在python中使用imaplib获取电子邮件正文?

时间:2010-02-09 15:10:15

标签: python imaplib

我想从IMAP4服务器获取整个消息。 在python docs中,如果发现这段代码有效:

>>> t, data = M.fetch('1', '(RFC822)')
>>> body = data[0][1]

我想知道我是否总能相信数据[0] [1]会返回消息正文。当我运行'RFC822.SIZE'时,我只有一个字符串而不是一个元组。

我已经浏览了rfc1730,但我无法找出'RFC822'的正确响应结构。从imaplib文档中判断获取结果结构也很困难。

以下是我在获取RFC822时所获得的内容:

('OK', [('1 (RFC822 {858569}', 'body of the message', ')')])

但是当我拿到RFC822.SIZE时,我得到了:

('OK', ['1 (RFC822.SIZE 847403)'])

我应该如何正确处理数据[0]列表? 我可以相信,当它是一个元组列表时,元组恰好有3个部分而第二部分是有效载荷吗?

也许你知道imap4的更好的库吗?

4 个答案:

答案 0 :(得分:28)

不...... imaplib是一个非常好的图书馆,它是如此难以理解的imap。

您可能希望检查t == 'OK',但data[0][1]按预期工作,就像我使用它一样。

这是我用来提取我通过电子邮件收到的签名证书的快速示例,而非防弹,但适合我的目的:

import getpass, os, imaplib, email
from OpenSSL.crypto import load_certificate, FILETYPE_PEM

def getMsgs(servername="myimapserverfqdn"):
  usernm = getpass.getuser()
  passwd = getpass.getpass()
  subject = 'Your SSL Certificate'
  conn = imaplib.IMAP4_SSL(servername)
  conn.login(usernm,passwd)
  conn.select('Inbox')
  typ, data = conn.search(None,'(UNSEEN SUBJECT "%s")' % subject)
  for num in data[0].split():
    typ, data = conn.fetch(num,'(RFC822)')
    msg = email.message_from_string(data[0][1])
    typ, data = conn.store(num,'-FLAGS','\\Seen')
    yield msg

def getAttachment(msg,check):
  for part in msg.walk():
    if part.get_content_type() == 'application/octet-stream':
      if check(part.get_filename()):
        return part.get_payload(decode=1)

if __name__ == '__main__':
  for msg in getMsgs():
    payload = getAttachment(msg,lambda x: x.endswith('.pem'))
    if not payload:
      continue
    try:
      cert = load_certificate(FILETYPE_PEM,payload)
    except:
      cert = None
    if cert:
      cn = cert.get_subject().commonName
      filename = "%s.pem" % cn
      if not os.path.exists(filename):
        open(filename,'w').write(payload)
        print "Writing to %s" % filename
      else:
        print "%s already exists" % filename

答案 1 :(得分:10)

IMAPClient包更易于使用。从描述:

  

易于使用,Pythonic和完整   IMAP客户端库。

答案 2 :(得分:1)

这是我提取有用信息的解决方案。到目前为止它一直很可靠:

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

答案 3 :(得分:0)

imap-tools 使用IMAP协议有效处理电子邮件的库。

  • 带字母属性的透明工作(包括uid)
  • 使用目录中的字母(复制,删除,标记,移动,看到)
  • 使用目录(列表,设置,获取,创建,存在,重命名,删除,状态)
  • 没有外部依赖