这段代码工作,并向我发送电子邮件就好:

import smtplib
#SERVER = "localhost"

FROM = 'monty@python.com'

TO = ["jon@mycompany.com"] # must be a list

SUBJECT = "Hello!"

TEXT = "This message was sent with Python's smtplib."

# Prepare actual message

message = """\
From: %s
To: %s
Subject: %s

%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)

# Send the mail

server = smtplib.SMTP('myserver')
server.sendmail(FROM, TO, message)
server.quit()

然而,如果我试图将它包装在这样一个函数中:

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    import smtplib
    """this is some test documentation in the function"""
    message = """\
        From: %s
        To: %s
        Subject: %s
        %s
        """ % (FROM, ", ".join(TO), SUBJECT, TEXT)
    # Send the mail
    server = smtplib.SMTP(SERVER)
    server.sendmail(FROM, TO, message)
    server.quit()

我得到以下错误:

 Traceback (most recent call last):
  File "C:/Python31/mailtest1.py", line 8, in <module>
    sendmail.sendMail(sender,recipients,subject,body,server)
  File "C:/Python31\sendmail.py", line 13, in sendMail
    server.sendmail(FROM, TO, message)
  File "C:\Python31\lib\smtplib.py", line 720, in sendmail
    self.rset()
  File "C:\Python31\lib\smtplib.py", line 444, in rset
    return self.docmd("rset")
  File "C:\Python31\lib\smtplib.py", line 368, in docmd
    return self.getreply()
  File "C:\Python31\lib\smtplib.py", line 345, in getreply
    raise SMTPServerDisconnected("Connection unexpectedly closed")
smtplib.SMTPServerDisconnected: Connection unexpectedly closed

有人能告诉我为什么吗?


当前回答

确保您已在电子邮件帐户中授予发件人和收件人发送和接收来自未知来源(外部来源)的电子邮件的权限。

import smtplib

#Ports 465 and 587 are intended for email client to email server communication - sending email
server = smtplib.SMTP('smtp.gmail.com', 587)

#starttls() is a way to take an existing insecure connection and upgrade it to a secure connection using SSL/TLS.
server.starttls()

#Next, log in to the server
server.login("#email", "#password")

msg = "Hello! This Message was sent by the help of Python"

#Send the mail
server.sendmail("#Sender", "#Reciever", msg)

其他回答

import smtplib, ssl

port = 587  # For starttls
smtp_server = "smtp.office365.com"
sender_email = "170111018@student.mit.edu.tr"
receiver_email = "professordave@hotmail.com"
password = "12345678"
message = """\
Subject: Final exam

Teacher when is the final exam?"""

def SendMailf():
    context = ssl.create_default_context()
    with smtplib.SMTP(smtp_server, port) as server:
        server.ehlo()  # Can be omitted
        server.starttls(context=context)
        server.ehlo()  # Can be omitted
        server.login(sender_email, password)
        server.sendmail(sender_email, receiver_email, message)
        print("mail send")

它可能会在你的信息中添加标签。在你把它传递给sendMail之前打印出消息。

我对发送电子邮件的包选项不满意,我决定制作并开源我自己的电子邮件发送器。它易于使用,并支持高级用例。

如何安装:

pip install redmail

用法:

from redmail import EmailSender
email = EmailSender(
    host="<SMTP HOST ADDRESS>",
    port=<PORT NUMBER>,
)

email.send(
    sender="me@example.com",
    receivers=["you@example.com"],
    subject="An example email",
    text="Hi, this is text body.",
    html="<h1>Hi,</h1><p>this is HTML body</p>"
)

如果您的服务器需要用户和密码,只需将user_name和密码传递给EmailSender。

我在send方法中包含了很多特性:

包含附件 将图像直接包含到HTML主体中 金贾的模板 漂亮的HTML表格开箱即用

文档: https://red-mail.readthedocs.io/en/latest/

源代码:https://github.com/Miksus/red-mail

在缩进函数中的代码时(这是可以的),还缩进了原始消息字符串的行。但是前导空白意味着标题行的折叠(连接),如RFC 2822 - Internet Message Format的2.2.3和3.2.3节所述:

每个报头字段在逻辑上是由一行字符组成的 字段名、冒号和字段主体。为了方便 但是,为了处理每行998/78个字符的限制, 报头字段的字段主体部分可以分成多个 线表示;这叫做“折叠”。

在sendmail调用的函数形式中,所有行都以空白开始,因此是“展开的”(连接),您正在尝试发送

From: monty@python.com    To: jon@mycompany.com    Subject: Hello!    This message was sent with Python's smtplib.

与我们的想法不同,smtplib将不再理解To:和Subject:头文件,因为这些名称只在一行的开头被识别。相反,smtplib将假设一个非常长的发送者电子邮件地址:

monty@python.com    To: jon@mycompany.com    Subject: Hello!    This message was sent with Python's smtplib.

这将不起作用,因此出现异常。

解决方案很简单:只保留原来的消息字符串。这可以通过一个函数来完成(正如Zeeshan建议的那样),也可以直接在源代码中完成:

import smtplib

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    """this is some test documentation in the function"""
    message = """\
From: %s
To: %s
Subject: %s

%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)
    # Send the mail
    server = smtplib.SMTP(SERVER)
    server.sendmail(FROM, TO, message)
    server.quit()

现在展开没有发生,你发送

From: monty@python.com
To: jon@mycompany.com
Subject: Hello!

This message was sent with Python's smtplib.

这就是您的旧代码所做的工作。

请注意,我还保留了标题和正文之间的空行,以适应RFC的第3.5节(这是必需的),并根据Python风格指南PEP-0008(这是可选的)将include放在函数之外。

值得注意的是,SMTP模块支持上下文管理器,因此不需要手动调用quit(),这将确保即使出现异常也始终调用它。

    with smtplib.SMTP_SSL('smtp.gmail.com', 465) as server:
        server.ehlo()
        server.login(user, password)
        server.sendmail(from, to, body)