下面的代码可以在python2中读取gmail收件箱。但是,即使我用括号括起了需要打印的内容,它也不能在python3中运行。
import smtplib
import time
import imaplib
import email
FROM_EMAIL = "email"
FROM_PWD = "pass"
SMTP_SERVER = "imap.gmail.com"
SMTP_PORT = 993
mail = imaplib.IMAP4_SSL(SMTP_SERVER)
mail.login(FROM_EMAIL,FROM_PWD)
mail.select('inbox')
type, data = mail.search(None, 'ALL')
mail_ids = data[0]
id_list = mail_ids.split()
first_email_id = int(id_list[0])
latest_email_id = int(id_list[-1])
for i in range(latest_email_id,first_email_id, -1):
typ, data = mail.fetch(i, '(RFC822)' )
for response_part in data:
if isinstance(response_part, tuple):
msg = email.message_from_string(response_part[1])
email_subject = msg['subject']
email_from = msg['from']
print('From : ' + email_from + '\n')
print('Subject : ' + email_subject + '\n')
它会给我一个错误:
data = data + b' ' + arg
TypeError: can't concat bytes to int
发布于 2018-06-05 01:50:41
有两个地方需要改变:
typ, data = mail.fetch(str(i), '(RFC822)' )
msg = email.message_from_string(response[1].decode())
希望这能有所帮助。
https://stackoverflow.com/questions/45788873
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