使用python发送邮件

Question

问题：当我向用户发送邮件时，从未在用户收件箱中看到的用户名仅显示电子邮件 ID，但我需要发件人的用户名

来自：demo@gmail.com 用户名：Demo

收件人：demotest@gmail.com

代码

import smtplib
fromaddr = From
toaddrs  = To
msg = 'Why,Oh why!'
username = From
password = *******
server = smtplib.SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()
server.login(username, password)
server.sendmail(fromaddr, toaddrs, msg)
server.quit()

Answer 1

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

fromaddr = 'demo@gmail.com'
toaddrs = 'demotest@gmail.com'

msg = MIMEMultipart('alternative')
msg['Subject'] = "Link"
msg['From'] = "good morning" #like name
msg['To'] = "GGGGGG"

body = MIMEText("example email body")
msg.attach(body)

username = 'demo@gmail.com'
password = ''
server = smtplib.SMTP_SSL('smtp.googlemail.com', 465)
server.login(username, password)
server.sendmail(fromaddr, toaddrs, msg.as_string())
server.quit()

Answer 2

或者走简单的路，安装yagmail和

给定：

To = 'someone@gmail.com'
From = 'me@gmail.com'
pwd = '******'
alias = 'someone'

运行：

import yagmail
yag = yagmail.SMTP(From, pwd)
yag.send({To: alias}, 'subject', 'Why,Oh why!')

安装可能由pip install yagmail完成

Answer 3

您只需要正确创建消息。我认为最方便的方法是使用特殊的消息对象。我放置了一个类，也许可以帮助您在项目中发送消息。

import os
import smtplib

from email.mime.text import MIMEText
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart



class EmailSender(object):
    def __init__(self, subject, to, config):
        self.__subject = subject
        self.__to = tuple(to) if hasattr(to, '__iter__') else (to,)

        self.__from = config['user']
        self.__password = config['password']
        self.__server = config['server']
        self.__port = config['port']

        self.__message = MIMEMultipart()
        self.__message['Subject'] = self.__subject
        self.__message['From'] = self.__from
        self.__message['To'] = ', '.join(self.__to)


    def add_text(self, text):
        self.__message.attach(
            MIMEText(text)
        )


    def add_image(self, img_path, name=None):
        if name is None:
            name = os.path.basename(img_path)

        with open(img_path, 'rb') as f:
            img_data = f.read()
            image = MIMEImage(img_data, name=name)
            self.__message.attach(image)


    def send(self):
        server = smtplib.SMTP_SSL(self.__server, self.__port)
        server.login(self.__from, self.__password)
        server.sendmail(self.__from, self.__to, self.__message.as_string())
        server.close()



sender = EmailSender("My letter", "my_target@email", {
    'user': "from@email",
    'password': "123456",
    'server': "mail.google.com"
    'port': 465
})
sender.add_text("Why,Oh why!")
sender.send()

Answer 4

smtplib 不会自动包含 any 标头，您需要包含 From: 标头，因此您必须自己放置一个，如下所示：

# Add the From: and To: headers at the start!
msg = ("From: %s\r\nTo: %s\r\n\r\n"
       % (fromaddr, ", ".join(toaddrs)))

您可以在DOCS 中看到。