"TypeError: can't concat int to bytes" imaplib error

Question

我正在创建一些东西，所以当它根据主题行收到一封电子邮件时，它会起作用，现在我的代码在 python 2.7 中工作并在 3.7 中运行，但给出一个错误“TypeError：无法将 int 连接到字节”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 的新手，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。所以字符串转换会修复它：

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

修复它不是必要的更改，但反转列表的更 pythonic 方法是使用列表切片 [::-1]。您可以保存一些行和 range() 调用。
此外你的缩进是错误的，你只处理最后一次迭代的数据。

...
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。）

"TypeError: can't concat int to bytes" imaplib error

"TypeError: can't concat int to bytes" imaplib error

python

byte

concatenation

imaplib