Python imaplib获取正文电子邮件gmail

Question

我已经读过这个并编写了这个脚本来获取某些邮箱中的电子邮件正文,标题以'$'开头,并由一些发件人发送.

import email, getpass, imaplib, os

detach_dir = "F:\PYTHONPROJECTS" # where you will save attachments
user = raw_input("Enter your GMail username --> ")
pwd = getpass.getpass("Enter your password --> ")

# connect to the gmail imap server
m = imaplib.IMAP4_SSL("imap.gmail.com")
m.login(user, pwd)

m.select("PETROLEUM") # here you a can choose a mail box like INBOX instead
# use m.list() to get all the mailboxes

resp, items = m.search(None, '(FROM "EIA_eLists@eia.gov")')
items = items[0].split() # getting the mails id

my_msg = [] # store relevant msgs here in please
msg_cnt = 0
break_ = False
for emailid in items[::-1]:
    resp, data = m.fetch(emailid, "(RFC822)")
    if ( break_ ):
        break 
    for response_part in data:
      if isinstance(response_part, tuple):
          msg = email.message_from_string(response_part[1])
          varSubject = msg['subject']
          if varSubject[0] == '$':
              msg_cnt += 1
              my_msg.append(msg)
              print msg_cnt
              print email.message_from_string(response_part[1])
              if ( msg_cnt == 5 ):
                  break_ = True 

如果我打印email.message_from_string(response_part[1]),我可以看到它包含第一个信息(标题,从,到,日期...),全文体.但是,我无法取得身体本身.email.message_from_string(response_part[0])打印邮件IDS,email.message_from_string(response_part[2])超出范围.email.message_from_string(response_part[1][0])也没有这样做.

感谢致敬.

UPDATE

现在,我几乎可以有正文.然而,它仍然被首先出现的信息声明所破坏.我得到了结果

From nobody Tue Dec 25 11:42:58 2012

US=3D$4.030

EastCst=3D$4.036

NewEng=3D$4.205

CenAtl=3D$4.149

LwrAtl=3D$3.921

Midwst=3D$3.984

GulfCst=3D$3.945

RkyMt=3D$4.195

WCst=3D$4.187

CA=3D$4.268

我想摆脱From nobody Tue Dec 25 11:42:58 2012哪些是信息.我知道我可以解析文本寻找第一条相关的线......我知道.

实现目的的代码(插入我的第一个样本)是

  if varSubject[0] == '$':
      r, d = m.fetch(emailid, "(UID BODY[TEXT])")
      msg_cnt += 1
      my_msg.append(msg)
      print email.message_from_string(d[0][1])

你有更好的方法(没有信息字符串)??? 更多:现在获取日期的命令是什么？我知道我可以varDate = msg['date']在上面适合的地方做,但是如何才能获得一个月的一年？谢谢

Answer 1

我已经设法使用Gmail工作,它提取有用的位并将它们输出到文本文件:

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

Answer 2

您可以通过执行以下任一操作来获取身体的内容

msg.as_string()
str(msg)
repr(msg)

http://docs.python.org/2.7/library/email.message.html#email.message.Message