python - 如何在 python 2.7 中从邮件的有效负载中仅打印消息正文

Question

我正在尝试打印消息，但收到不需要的文本。无法过滤。

#!/usr/bin/python
import imaplib
import email
import re
p = re.compile(r'Server Status')

mail = imaplib.IMAP4_SSL('stbeehive.yxz.com')
(retcode, capabilities) = mail.login('abc@yxz.com','passwd')
print retcode, capabilities
mail.list()
mail.select('Inbox')
n=0
(retcode, messages) = mail.search(None,'(UNSEEN)')
if retcode == 'OK':
    for num in messages[0].split() :
        print 'Processing '
        n=n+1
        typ, data = mail.fetch(num,'(RFC822)')
        for response_part in data:
             if isinstance(response_part, tuple):
                 original = email.message_from_string(response_part[1])
                 print original['From']
                 print original['Subject']
                 if original.is_multipart():
                     message =  original.get_payload()[0]
                     print message
                     for line in message:
                         if p.findall(line):
                             print line
                 else:
                     print original.get_payload()

print n

当我尝试打印下面的消息时。我只想要其中的第三行。

Content-Type: text/plain; charset=us-ascii
Content-Transfer-Encoding: quoted-printable
Server Status#XYZBSS##XYZ Running

我尝试在上面的代码中使用 re.compile 来过滤“服务器状态”，但出现以下错误。

  File "./mail.py", line 27, in <module>
    for line in message:
  File "/usr/lib64/python2.6/email/message.py", line 292, in __getitem__
    return self.get(name)
  File "/usr/lib64/python2.6/email/message.py", line 358, in get
    name = name.lower()
AttributeError: 'int' object has no attribute 'lower'

Answer 1

如果没有看到该消息，这有点推测性，但看起来您正在提取带有标题和所有内容的正文部分。您想要找到正确的 body 部位，然后提取其有效负载。

如果没有实际的消息可供查看，则无法对此进行测试，但我猜测类似

             if original.is_multipart():
                 # Quick hack, should probably properly recurse
                 message =  original.get_payload()[0].get_payload()
             else:
                 message = original.get_payload()
             #print message
             for line in message.split('\n'):
                 if 'Server Status' in line:   # look ma, no regex necessary
                     print line