3.3不是2.7的poplib错误:无效的消息号

Question

我从Python 3.3上使用poplib的代码中获得错误，但它适用于Python 2.7:

poplib.error_proto: b"-ERR Invalid message number: b'1'"

我想迁移到python 3.3，因为我有一个特定的模块只安装在我的python 3.3上。

我正在学习python编程语言。

下面是在python 2.7上成功的示例，但是这个示例代码不能在我的python 3.3上工作。

import poplib
pop_server = 'mail01.org'
user = 'user'
password = 'pass'
p = poplib.POP3(pop_server)
p.user(user)
p.pass_(password)
print ("This mailbox has %d messages, totaling %d bytes." % p.stat())
msg_list = p.list()
print (msg_list)
for msg in msg_list[1]:
    msg_num, _ = msg.split()
    resp = p.retr(msg_num)

输出如下:

This mailbox has 2 messages, totaling 633300 bytes.
(b'+OK 2 messages:', [b'1 137956', b'2 495344'], 20)

下面是错误回溯:

Traceback (most recent call last):
  File "AttachmentDownloader.py", line 28, in <module>
    resp = p.retr(msg_num)
  File "C:Python33libpoplib.py", line 236, in retr
    return self._longcmd('RETR %s' % which)
  File "C:Python33libpoplib.py", line 171, in _longcmd
    return self._getlongresp()
  File "C:Python33libpoplib.py", line 147, in _getlongresp
    resp = self._getresp()
  File "C:Python33libpoplib.py", line 140, in _getresp
    raise error_proto(resp)
poplib.error_proto: b"-ERR Invalid message number: b'1'"

Answer 1

您正在尝试传递str作为消息号码。更改以下行

msg_num, _ = msg.split()

msg_num = int(msg.split()[0])