我read this already并编写此脚本以获取某些邮箱中标题以“$”开头并由某些发件人发送的电子邮件正文。

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、发件人、收件人、日期...)，全文正文。但是，我无法获取 body 本身。 email.message_from_string(response_part[0]) 打印邮件 IDS，而 email.message_from_string(response_part[2]) 超出范围。 email.message_from_string(response_part[1][0]) 两者都没有这样做。

感谢和问候。

更新

现在，我几乎可以拥有正文了。然而，它仍然被首先出现的信息声明所破坏。结果我得到

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 这是信息。我知道我可以解析文本寻找第一个相关行...我知道。

实现此目的的代码(插入我的第一个示例)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])

你有更好的方法吗(没有信息字符串)？？？更多:现在获取日期的命令是什么？我知道我可以在上面适合的地方执行 varDate = msg['date'] ，但是如何获取 day-month-year ？谢谢

最佳答案

我已经设法使用 Gmail 让它工作，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

关于Python imaplib 获取正文电子邮件 gmail，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/14029768/