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")

有压痕问题。下面的代码将工作:

import textwrap

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    import smtplib
    """this is some test documentation in the function"""
    message = textwrap.dedent("""\
        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()

我编写了一个简单的函数send_email()，用于使用smtplib和电子邮件包发送电子邮件(链接到我的文章)。它还使用dotenv包来加载发件人的电子邮件和密码(请不要在代码中保密!)我正在使用Gmail电子邮件服务。密码是应用程序密码(这里是谷歌文档如何生成应用程序密码)。

import os
import smtplib
from email.message import EmailMessage
from dotenv import load_dotenv
_ = load_dotenv()


def send_email(to, subject, message):
    try:
        email_address = os.environ.get("EMAIL_ADDRESS")
        email_password = os.environ.get("EMAIL_PASSWORD")

        if email_address is None or email_password is None:
            # no email address or password
            # something is not configured properly
            print("Did you set email address and password correctly?")
            return False

        # create email
        msg = EmailMessage()
        msg['Subject'] = subject
        msg['From'] = email_address
        msg['To'] = to
        msg.set_content(message)

        # send email
        with smtplib.SMTP_SSL('smtp.gmail.com', 465) as smtp:
            smtp.login(email_address, email_password)
            smtp.send_message(msg)
        return True
    except Exception as e:
        print("Problem during send email")
        print(str(e))
    return False

以上方法对于简单的电子邮件发送是可行的。如果您正在寻找更高级的功能，例如HTML内容或附件，当然可以手工编码，但我建议使用现有的包，例如yagmail。

Gmail每天限制500封邮件。对于每天发送许多电子邮件，请考虑事务性电子邮件服务提供商，如Amazon SES, MailGun, MailJet或SendGrid。

我想通过建议yagmail包来帮助你发送电子邮件(我是维护者，抱歉广告，但我觉得它真的能帮助!)

你的整个代码将是:

import yagmail
yag = yagmail.SMTP(FROM, 'pass')
yag.send(TO, SUBJECT, TEXT)

注意，我为所有参数提供了默认值，例如，如果你想发送给自己，你可以省略to，如果你不想要一个主题，你也可以省略它。

此外，我们的目标还在于使附加html代码或图像(以及其他文件)变得非常容易。

在你放置内容的地方，你可以这样做:

contents = ['Body text, and here is an embedded image:', 'http://somedomain/image.png',
            'You can also find an audio file attached.', '/local/path/song.mp3']

哇，发送附件是多么简单啊!如果没有yagmail，这大概需要20行;)

此外，如果你设置了一次，你就永远不必再输入密码(并安全地保存密码)。在你的情况下，你可以这样做:

import yagmail
yagmail.SMTP().send(contents = contents)

这样更简洁!

我建议你看看github，或者直接用pip install yagmail安装它。

我建议您使用标准包email和smtplib一起发送电子邮件。请看下面的例子(摘自Python文档)。注意，如果遵循这种方法，“简单”任务确实很简单，而更复杂的任务(如附加二进制对象或发送纯/HTML多部分消息)将很快完成。

# Import smtplib for the actual sending function
import smtplib

# Import the email modules we'll need
from email.mime.text import MIMEText

# Open a plain text file for reading.  For this example, assume that
# the text file contains only ASCII characters.
with open(textfile, 'rb') as fp:
    # Create a text/plain message
    msg = MIMEText(fp.read())

# me == the sender's email address
# you == the recipient's email address
msg['Subject'] = 'The contents of %s' % textfile
msg['From'] = me
msg['To'] = you

# Send the message via our own SMTP server, but don't include the
# envelope header.
s = smtplib.SMTP('localhost')
s.sendmail(me, [you], msg.as_string())
s.quit()

要将电子邮件发送到多个目的地，您还可以遵循Python文档中的示例:

# Import smtplib for the actual sending function
import smtplib

# Here are the email package modules we'll need
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart

# Create the container (outer) email message.
msg = MIMEMultipart()
msg['Subject'] = 'Our family reunion'
# me == the sender's email address
# family = the list of all recipients' email addresses
msg['From'] = me
msg['To'] = ', '.join(family)
msg.preamble = 'Our family reunion'

# Assume we know that the image files are all in PNG format
for file in pngfiles:
    # Open the files in binary mode.  Let the MIMEImage class automatically
    # guess the specific image type.
    with open(file, 'rb') as fp:
        img = MIMEImage(fp.read())
    msg.attach(img)

# Send the email via our own SMTP server.
s = smtplib.SMTP('localhost')
s.sendmail(me, family, msg.as_string())
s.quit()

如您所见，MIMEText对象中的报头To必须是由逗号分隔的电子邮件地址组成的字符串。另一方面，sendmail函数的第二个参数必须是一个字符串列表(每个字符串都是一个电子邮件地址)。

所以，如果你有三个电子邮件地址:person1@example.com, person2@example.com和person3@example.com，你可以这样做(明显的部分省略了):

to = ["person1@example.com", "person2@example.com", "person3@example.com"]
msg['To'] = ",".join(to)
s.sendmail(me, to, msg.as_string())

"，".join(to)部分从列表中生成一个单独的字符串，用逗号分隔。

从你的问题中，我猜你还没有读过Python教程——如果你想在Python中有所了解，这是必须的——标准库的文档大部分都很出色。

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")