“TypeError：无法将 int 连接到字节” imaplib 错误

Question

我正在创建一些东西，所以当它收到一封取决于主题行的电子邮件时，它现在执行一个函数，我的代码在 python 2.7 中工作并在 3.7 中运行，但给出一个错误“TypeError: can't concat int to bytes" 我没有尝试太多，因为我在网上找不到任何东西，而且我是 python 新手，请告诉我

import imaplib
import email
from time import sleep
from ssh import myRoomOff, myRoomOn

count = 0 

mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login('EMAIL', 'PASSWORD')
mail.list()
mail.select('inbox')

#need to add some stuff in here
while True:
    mail.select('inbox')
    typ, data = mail.search(None, 'UNSEEN')
    ids = data[0]
    id_list = ids.split()


    #get the most recent email id
    latest_email_id = int( id_list[-1] )

    #iterate through 15 messages in decending order starting with latest_email_id
    #the '-1' dictates reverse looping order
    for i in range( latest_email_id, latest_email_id-1, -1 ):
        typ, data = mail.fetch(i, '(RFC822)' )

    for response_part in data:
        if isinstance(response_part, tuple):
            msg = email.message_from_bytes(response_part[1])
            varSubject = msg['subject']

            count += 1
            print(str(count) + ", " + varSubject)
            if varSubject == "+myRoom":
                myRoomOn()
            elif varSubject == "-myRoom":
                myRoomOff()
            else:
                print("I do not understand this email!")
                pass
    sleep(2)

错误

Traceback (most recent call last):
  File "/Users/danielcaminero/Desktop/alexaCommandThing/checkForEmail.py", line 28, in <module>
    typ, data = mail.fetch(i, '(RFC822)' )
  File "/Library/Frameworks/Python.framework/Versions/3.9/lib/python3.9/imaplib.py", line 548, in fetch
    typ, dat = self._simple_command(name, message_set, message_parts)
  File "/Library/Frameworks/Python.framework/Versions/3.9/lib/python3.9/imaplib.py", line 1230, in _simple_command
    return self._command_complete(name, self._command(name, *args))
  File "/Library/Frameworks/Python.framework/Versions/3.9/lib/python3.9/imaplib.py", line 988, in _command
    data = data + b' ' + arg
TypeError: can't concat int to bytes

Answer 1

mail.fetch() 的第一个参数必须是字符串，但range() 产生一个 int。 所以字符串转换可以解决它：

typ, data = mail.fetch(str(i), '(RFC822)')

---

修复它不是必要的更改，但是一种更 Pythonic 的反转列表的方法是使用列表切片[::-1]。您可以保存一些行和range() 调用。
此外，您的缩进是错误的，您只处理上次迭代的数据。

...
id_list = ids.split()

for i in id_list[::-1]:
    typ, data = mail.fetch(i, '(RFC822)')
    for response_part in data:
        if isinstance(response_part, tuple):
            msg = email.message_from_bytes(response_part[1])
            varSubject = msg['subject']

            count += 1
            print(str(count) + ", " + varSubject)
            if varSubject == "+myRoom":
                myRoomOn()
            elif varSubject == "-myRoom":
                myRoomOff()
            else:
                print("I do not understand this email!")
                pass

（这里不需要字符串转换，因为id_list 是一个字节串列表。字节串也是fetch() 的有效值。）