我无法看到我在哪里出错,我希望有人能发现问题。我想发一封电子邮件到多个地址;但是,它只将它发送到列表中的第一个电子邮件地址,而不是两者。这是代码:
import smtplib
from smtplib import SMTP
recipients = ['example1@gmail.com', 'example2@example.com']
def send_email (message, status):
fromaddr = 'from@gmail.com'
toaddrs = ", ".join(recipients)
server = SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()
server.ehlo()
server.login('example_username', 'example_pw')
server.sendmail(fromaddr, toaddrs, 'Subject: %s\r\n%s' % (status, message))
server.quit()
send_email("message","subject")
之前有没有人遇到此错误?
感谢您的时间。
答案 0 :(得分:9)
尝试使用此代码,不使用您的加入:
import smtplib
from smtplib import SMTP
recipients = ['example1@gmail.com', 'example2@example.com']
def send_email (message, status):
fromaddr = 'from@gmail.com'
server = SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()
server.ehlo()
server.login('example_username', 'example_pw')
server.sendmail(fromaddr, recipients, 'Subject: %s\r\n%s' % (status, message))
server.quit()
send_email("message","subject")
希望它有所帮助!
答案 1 :(得分:6)
import smtplib
from email.mime.text import MIMEText
s = smtplib.SMTP('xxx.xx')
msg = MIMEText("""body""")
sender = 'xx.xx.com'
recipients = ['example1@gmail.com', 'example2@example.com']
msg['Subject'] = "test"
msg['From'] = sender
msg['To'] = ", ".join(recipients)
s.sendmail(sender, recipients, msg.as_string())
答案 2 :(得分:5)
更改
toaddrs = ", ".join(recipients)
到
toaddrs = recipients
自
server.sendmail(fromaddr, toaddrs, ...)
希望toaddrs
成为电子邮件地址的列表。 (或者,当然,只需使用recipients
代替toaddrs
。)