使用python发送邮件
问题:当我向用户发送邮件时,从用户收件箱中看不到的用户名到用户收件箱只显示电子邮件id,但我需要发件人的用户名 发件人:demo@gmail.com用户名:演示 致:demotest@gmail.com 代码使用python发送邮件,python,python-2.7,email,smtplib,Python,Python 2.7,Email,Smtplib,问题:当我向用户发送邮件时,从用户收件箱中看不到的用户名到用户收件箱只显示电子邮件id,但我需要发件人的用户名 发件人:demo@gmail.com用户名:演示 致:demotest@gmail.com 代码 smtplib不会自动包含任何标题,您需要包含一个From:标题,因此您必须自己放置一个标题,执行如下操作: # Add the From: and To: headers at the start! msg = ("From: %s\r\nTo: %s\r\n\r\n" %
smtplib
不会自动包含任何标题,您需要包含一个From:
标题,因此您必须自己放置一个标题,执行如下操作:
# Add the From: and To: headers at the start!
msg = ("From: %s\r\nTo: %s\r\n\r\n"
% (fromaddr, ", ".join(toaddrs)))
正如您可以在中使用一样。您只需要正确创建消息。我认为最方便的方法是使用一个特殊的对象来传递消息。我放置了一个类,它可能会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()
或者简单地安装和安装 鉴于:
To = 'someone@gmail.com'
From = 'me@gmail.com'
pwd = '******'
alias = 'someone'
运行:
安装可以通过
pip Install yagmail
完成,可能是Okay的副本,但问题可能是头部,让我们继续调查。如果您找到了解决方案,请告知我们获得了类似回溯(最近一次呼叫)的错误:socket.error:[Errno 101]无法访问网络可能会,这将帮助您:谢谢,但不是必需的yagmail LIB
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()
import yagmail
yag = yagmail.SMTP(From, pwd)
yag.send({To: alias}, 'subject', 'Why,Oh why!')
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()