在您的gmail帐户上启用不太安全的应用程序，并使用(Python>=3.6):

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

gmailUser = 'XXXXX@gmail.com'
gmailPassword = 'XXXXX'
recipient = 'XXXXX@gmail.com'

message = f"""
Type your message here...
"""

msg = MIMEMultipart()
msg['From'] = f'"Your Name" <{gmailUser}>'
msg['To'] = recipient
msg['Subject'] = "Subject here..."
msg.attach(MIMEText(message))

try:
    mailServer = smtplib.SMTP('smtp.gmail.com', 587)
    mailServer.ehlo()
    mailServer.starttls()
    mailServer.ehlo()
    mailServer.login(gmailUser, gmailPassword)
    mailServer.sendmail(gmailUser, recipient, msg.as_string())
    mailServer.close()
    print ('Email sent!')
except:
    print ('Something went wrong...')

你同意OOP吗?

#!/usr/bin/env python


import smtplib

class Gmail(object):
    def __init__(self, email, password):
        self.email = email
        self.password = password
        self.server = 'smtp.gmail.com'
        self.port = 587
        session = smtplib.SMTP(self.server, self.port)        
        session.ehlo()
        session.starttls()
        session.ehlo
        session.login(self.email, self.password)
        self.session = session

    def send_message(self, subject, body):
        ''' This must be removed '''
        headers = [
            "From: " + self.email,
            "Subject: " + subject,
            "To: " + self.email,
            "MIME-Version: 1.0",
           "Content-Type: text/html"]
        headers = "\r\n".join(headers)
        self.session.sendmail(
            self.email,
            self.email,
            headers + "\r\n\r\n" + body)


gm = Gmail('Your Email', 'Password')

gm.send_message('Subject', 'Message')

不直接相关，但仍然值得指出的是，我的包试图使发送gmail消息非常快速和轻松。它还试图维护一个错误列表，并试图立即指出解决方案。

它只需要下面的代码来完成你所写的:

import yagmail
yag = yagmail.SMTP('user_me@gmail.com')
yag.send('user_you@gmail.com', 'Why,Oh why!')

或者一句话:

yagmail.SMTP('user_me@gmail.com').send('user_you@gmail.com', 'Why,Oh why!')

关于包/安装，请查看git或pip，可用于Python 2和3。

    import smtplib

    fromadd='from@gmail.com'
    toadd='send@gmail.com'

    msg='''hi,how r u'''
    username='abc@gmail.com'
    passwd='password'

    try:
        server = smtplib.SMTP('smtp.gmail.com:587')
        server.ehlo()
        server.starttls()
        server.login(username,passwd)

        server.sendmail(fromadd,toadd,msg)
        print("Mail Send Successfully")
        server.quit()

   except:
        print("Error:unable to send mail")

   NOTE:https://www.google.com/settings/security/lesssecureapps that                                                         should be enabled

def send_email(user, pwd, recipient, subject, body):
    import smtplib

    FROM = user
    TO = recipient if isinstance(recipient, list) else [recipient]
    SUBJECT = subject
    TEXT = body

    # Prepare actual message
    message = """From: %s\nTo: %s\nSubject: %s\n\n%s
    """ % (FROM, ", ".join(TO), SUBJECT, TEXT)
    try:
        server = smtplib.SMTP("smtp.gmail.com", 587)
        server.ehlo()
        server.starttls()
        server.login(user, pwd)
        server.sendmail(FROM, TO, message)
        server.close()
        print 'successfully sent the mail'
    except:
        print "failed to send mail"

如果你想使用端口465，你必须创建一个SMTP_SSL对象:

# SMTP_SSL Example
server_ssl = smtplib.SMTP_SSL("smtp.gmail.com", 465)
server_ssl.ehlo() # optional, called by login()
server_ssl.login(gmail_user, gmail_pwd)  
# ssl server doesn't support or need tls, so don't call server_ssl.starttls() 
server_ssl.sendmail(FROM, TO, message)
#server_ssl.quit()
server_ssl.close()
print 'successfully sent the mail'

你需要在直接运行到STARTTLS之前说EHLO:

server = smtplib.SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()

此外，您应该真正创建From:、To:和Subject:消息头，用空行与消息体分隔，并使用CRLF作为EOL标记。

E.g.

msg = "\r\n".join([
  "From: user_me@gmail.com",
  "To: user_you@gmail.com",
  "Subject: Just a message",
  "",
  "Why, oh why"
  ])

注意:

为了做到这一点，你需要在gmail帐户配置中启用“允许不太安全的应用程序”选项。否则，当gmail检测到一个非谷歌应用程序试图登录你的帐户时，你会得到一个“关键安全警报”。