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python - 使用 imaplib 从特定收件箱接收电子邮件时出错

转载 作者:太空宇宙 更新时间:2023-11-03 21:25:20 24 4
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从特定文件夹 (Office365) 读取电子邮件时,邮件正文是无意义的随机长字符串。

最初,我将这些电子邮件作为副本发送到我的个人邮件中的特定文件夹中。从那里阅读我没有遇到任何问题。但是当我尝试直接从主收件箱读取时,消息是没有意义的长字符串(因此,我无法解析任何内容)

mail = imaplib.IMAP4_SSL(SMTP_SERVER)
mail.login(FROM_EMAIL, FROM_PWD)
boxes = mail.list()
mail.select('INBOX/netstatus', readonly=True)
(type, data) = mail.uid('SEARCH', None, '(UNSEEN)')
mail_ids = data[0]
id_list = mail_ids.split()


def read_email(self, *id_list):
id_list = list(id_list)
for i in reversed(id_list):
(typ, data) = mail.uid('FETCH', i, '(RFC822)')
for response_part in data:
if isinstance(response_part, tuple):
print("Response: ",response_part)
msg = \ email.message_from_string(response_part[1].decode('utf-8'))

body = ''
email_subject = msg['subject']
email_from = msg['from']
email_date = msg['date']
message= msg.get_payload().encode('utf-8')
print(message)

当我从我的个人收件箱中阅读时,我收到类似的内容:

mymail: b'Status from Device (x.x.x.x) to AnotherDevice (y.y.y.y), interface A B, in Protocol came up\r\n'

但是当我从主收件箱中阅读时:

'QWRq4N5IGZyb20gVVREFxLXJlMSAuMTAuMjUsIDAxMDAuMTAwMS4w\r\nMDI1LjAwKSB0byBVU0FEQUxIERS4MC4xMC4xMMDAxLjAw\r\nMjUuMDQpLCBpbnZhY2UgMTAuMTA4AuMjI5IHRvIDEwLjEwLjIwLjEzMCwgaW4gSVNJUy9M\r\nZXZlbDIgd2VudCBkkNvbmZ3VyZWQgd2F0Y2hsaXN0OiI1OQ0K\r\n'

最佳答案

使用Python3.6或更高版本,您可以使用EmailMessage.get_content()自动解码编码内容。

import email
from email import policy

# Email based on one of the examples at https://docs.python.org/3.6/library/email.examples.html

s = """Subject: Ayons asperges pour le =?utf-8?q?d=C3=A9jeuner?=
From: =?utf-8?q?Pep=C3=A9?= Le Pew <pepe@example.com>
To: Penelope Pussycat <penelope@example.com>,
Fabrette Pussycat <fabrette@example.com>
Content-Type: text/plain; charset="utf-8"
Content-Transfer-Encoding: base64
MIME-Version: 1.0

ICAgIFNhbHV0IQoKICAgIENlbGEgcmVzc2VtYmxlIMOgIHVuIGV4Y2VsbGVudCByZWNpcGllWzFd
IGTDqWpldW5lci4KCiAgICBbMV0gaHR0cDovL3d3dy55dW1tbHkuY29tL3JlY2lwZS9Sb2FzdGVk
LUFzcGFyYWd1cy1FcGljdXJpb3VzLTIwMzcxOAoKICAgIC0tUGVww6kKICAgIAo=
"""

msg = email.message_from_string(s, policy=policy.default)
print(msg.get_content())

结果:

    Salut!

Cela ressemble à un excellent recipie[1] déjeuner.

[1] http://www.yummly.com/recipe/Roasted-Asparagus-Epicurious-203718

--Pepé

对于多部分消息,您需要迭代各部分:

for part in msg.iter_parts():
print(part.get_content())

关于python - 使用 imaplib 从特定收件箱接收电子邮件时出错,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53877899/

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