Python imaplib抓取正文電子郵件gmail

Question

我read this already並寫了這個腳本來獲取某些郵箱中的電子郵件正文，其標題以'$'開頭，並由某個發件人發送。

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信息（標題，發件人，收件人，日期等），則全文身體。但是，我無法獲取身體本身。 email.message_from_string(response_part[0])打印郵件IDS，並且email.message_from_string(response_part[2])超出範圍。 email.message_from_string(response_part[1][0])都沒有這樣做。

感謝和問候。

UPDATE

現在，我幾乎可以擁有正文文本。然而，它仍然首先出現在信息聲明中。我得到的結果

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 

我想擺脫From nobody Tue Dec 25 11:42:58 2012這是信息。我知道我可以解析文本尋找第一個相關的線......我知道。

實現如此（我的第一個樣品中塞）的代碼是

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]) 

你有一個更好的方法（無信息字符串）???更多：現在獲取日期的命令是什麼？我知道我可以在varDate = msg['date']以上適合，但是如何取得日日月？感謝

Answer 1

您可以通過執行以下任一

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

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

Answer 2

我已經成功地得到這個使用Gmail的工作得到了正文的內容，它提取有用位和輸出他們到文本文件：

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