经过多次搜索,我找不到如何使用smtplib.sendmail向多个收件人发送邮件。问题是,每次发送邮件时,邮件标题似乎都包含多个地址,但实际上只有第一个收件人会收到电子邮件。
问题似乎是email.Message
模块期望与smtplib.sendmail()
函数不同的东西。
简而言之,要发送给多个收件人,您应该将邮件头设置为逗号分隔的电子邮件地址字符串。但是,sendmail()
参数to_addrs
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()
发布于 2015-01-29 06:43:54
msg['To']
需要是一个字符串:
msg['To'] = "a@b.com, b@b.com, c@b.com"
而sendmail(sender, recipients, message)
中的recipients
需要是一个列表:
sendmail("a@a.com", ["a@b.com", "b@b.com", "c@b.com"], "Howdy")
发布于 2016-07-07 15:37:20
这对我很管用。
import smtplib
from email.mime.text import MIMEText
s = smtplib.SMTP('smtp.uk.xensource.com')
s.set_debuglevel(1)
msg = MIMEText("""body""")
sender = 'me@example.com'
recipients = 'john.doe@example.com,john.smith@example.co.uk'
msg['Subject'] = "subject line"
msg['From'] = sender
msg['To'] = recipients
s.sendmail(sender, recipients.split(','), msg.as_string())
发布于 2019-02-01 19:35:51
下面的解决方案对我很有效。它成功地向多个收件人发送电子邮件,包括“抄送”和“密件抄送”。
toaddr = ['mailid_1','mailid_2']
cc = ['mailid_3','mailid_4']
bcc = ['mailid_5','mailid_6']
subject = 'Email from Python Code'
fromaddr = 'sender_mailid'
message = "\n !! Hello... !!"
msg['From'] = fromaddr
msg['To'] = ', '.join(toaddr)
msg['Cc'] = ', '.join(cc)
msg['Bcc'] = ', '.join(bcc)
msg['Subject'] = subject
s.sendmail(fromaddr, (toaddr+cc+bcc) , message)
https://stackoverflow.com/questions/8856117
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