经过大量的搜索,我无法找到如何使用smtplib。发送邮件到多个收件人。问题是每次发送邮件时,邮件标题似乎包含多个地址,但实际上只有第一个收件人会收到电子邮件。
问题似乎出在邮件上。Message模块期望与smtplb .sendmail()函数不同的内容。
简而言之,要发送给多个收件人,您应该将标题设置为一串以逗号分隔的电子邮件地址。sendmail()参数to_addr应该是一个电子邮件地址列表。
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
import smtplib
msg = MIMEMultipart()
msg["Subject"] = "Example"
msg["From"] = "me@example.com"
msg["To"] = "malcom@example.com,reynolds@example.com,firefly@example.com"
msg["Cc"] = "serenity@example.com,inara@example.com"
body = MIMEText("example email body")
msg.attach(body)
smtp = smtplib.SMTP("mailhost.example.com", 25)
smtp.sendmail(msg["From"], msg["To"].split(",") + msg["Cc"].split(","), msg.as_string())
smtp.quit()
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
def sender(recipients):
body = 'Your email content here'
msg = MIMEMultipart()
msg['Subject'] = 'Email Subject'
msg['From'] = 'your.email@gmail.com'
msg['To'] = (', ').join(recipients.split(','))
msg.attach(MIMEText(body,'plain'))
server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login('your.email@gmail.com', 'yourpassword')
server.send_message(msg)
server.quit()
if __name__ == '__main__':
sender('email_1@domain.com,email_2@domain.com')
它只适用于我send_message函数和使用列表中的连接函数与收件人,python 3.6。
这里有很多答案在技术上或部分上是正确的。在阅读了每个人的答案后,我想出了这个更可靠/通用的电子邮件功能。我已经确认它的工作,你可以通过HTML或纯文本的主体。注意,此代码不包括附件代码:
import smtplib
import socket
# Import the email modules we'll need
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
#
# @param [String] email_list
# @param [String] subject_line
# @param [String] error_message
def sendEmailAlert(email_list="default@email.com", subject_line="Default Subject", error_message="Default Error Message"):
hostname = socket.gethostname()
# Create message
msg = MIMEMultipart()
msg['Subject'] = subject_line
msg['From'] = f'no-reply@{hostname}'
msg['To'] = email_list
msg.attach(MIMEText(error_message, 'html'))
# Send the message via SMTP server
s = smtplib.SMTP('localhost') # Change for remote mail server!
# Verbose debugging
s.set_debuglevel(2)
try:
s.sendmail(msg['From'], msg['To'].split(","), msg.as_string())
except Exception as e:
print(f'EMAIL ISSUE: {e}')
s.quit()
这显然可以修改为使用本机Python日志记录。我只是提供了一个坚实的核心功能。我也不能强调这一点,sendmail()需要一个列表,而不是一个字符串!函数适用于Python3.6+
我提出了这个可导入模块函数。本例中使用的邮件服务器为gmail。它分为头和消息,所以你可以清楚地看到发生了什么:
import smtplib
def send_alert(subject=""):
to = ['email@one.com', 'email2@another_email.com', 'a3rd@email.com']
gmail_user = 'me@gmail.com'
gmail_pwd = 'my_pass'
smtpserver = smtplib.SMTP("smtp.gmail.com", 587)
smtpserver.ehlo()
smtpserver.starttls()
smtpserver.ehlo
smtpserver.login(gmail_user, gmail_pwd)
header = 'To:' + ", ".join(to) + '\n' + 'From: ' + gmail_user + '\n' + 'Subject: ' + subject + '\n'
msg = header + '\n' + subject + '\n\n'
smtpserver.sendmail(gmail_user, to, msg)
smtpserver.close()
尝试声明一个包含所有收件人和cc_收件人的列表变量为字符串,而不是循环遍历它们,如下所示:
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
import smtplib
recipients = ["malcom@example.com","reynolds@example.com", "firefly@example.com"]
cc_recipients=["serenity@example.com", "inara@example.com"]
msg = MIMEMultipart()
msg["Subject"] = "Example"
msg["From"] = "me@example.com"
msg["To"] = ', '.join(recipients)
msg["Cc"] = ', '.join(cc_recipients)
body = MIMEText("example email body")
msg.attach(body)
smtp = smtplib.SMTP("mailhost.example.com", 25)
for recipient in recipients:
smtp.sendmail(msg["From"], recipient, msg.as_string())
for cc_recipient in cc_recipients:
smtp.sendmail(msg["From"], cc_recipient, msg.as_string())
smtp.quit()
实际上问题在于SMTP。发送邮件和电子邮件。MIMEText需要两个不同的东西。
电子邮件。MIMEText为电子邮件正文设置了“To:”标头。它仅用于向另一端的人显示结果,并且像所有电子邮件标题一样,必须是单个字符串。(请注意,它实际上不必与实际接收消息的人有任何关系。)
SMTP。另一方面,sendmail为SMTP协议设置消息的“信封”。它需要一个Python字符串列表,每个字符串都有一个地址。
所以,你需要做的就是将收到的两个回复结合起来。将msg['To']设置为单个字符串,但将原始列表传递给sendmail:
emails = ['a.com','b.com', 'c.com']
msg['To'] = ', '.join( emails )
....
s.sendmail( msg['From'], emails, msg.as_string())
要向多个收件人发送电子邮件,请将收件人添加为电子邮件id列表。
接收者= ['user1@email.com', 'user2@email.com', 'user3@email.com']
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
smtp_server = 'smtp-example.com'
port = 26
sender = 'user@email.com'
debuglevel = 0
# add receivers as list of email id string
receivers = ['user1@email.com', 'user2@email.com', 'user3@email.com']
message = MIMEMultipart(
"mixed", None, [MIMEImage(img_data, 'png'), MIMEText(html,'html')])
message['Subject'] = "Token Data"
message['From'] = sender
message['To'] = ", ".join(receivers)
try:
server = smtplib.SMTP('smtp-example.com')
server.set_debuglevel(1)
server.sendmail(sender, receivers, message.as_string())
server.quit()
# print(response)
except BaseException:
print('Error: unable to send email')