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()
对于那些希望只发送一个“to”报头的消息，下面的代码可以解决这个问题。确保你的接收者变量是一个字符串列表。
# Create message container - the correct MIME type is multipart/alternative.
msg = MIMEMultipart('alternative')
msg['Subject'] = title
msg['From'] = f'support@{config("domain_base")}'
msg['To'] = "me"
message_content += f"""
<br /><br />
Regards,<br />
Company Name<br />
The {config("domain_base")} team
"""
body = MIMEText(message_content, 'html')
msg.attach(body)
try:
smtpObj = smtplib.SMTP('localhost')
for r in receivers:
del msg['To']
msg['To'] = r #"Customer /n" + r
smtpObj.sendmail(f"support@{config('domain_base')}", r, msg.as_string())
smtpObj.quit()
return {"message": "Successfully sent email"}
except smtplib.SMTPException:
return {"message": "Error: unable to send email"}
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)
您可以在将收件人电子邮件写入文本文件时尝试此操作
from email.mime.text import MIMEText
from email.header import Header
import smtplib
f = open('emails.txt', 'r').readlines()
for n in f:
emails = n.rstrip()
server = smtplib.SMTP('smtp.uk.xensource.com')
server.ehlo()
server.starttls()
body = "Test Email"
subject = "Test"
from = "me@example.com"
to = emails
msg = MIMEText(body,'plain','utf-8')
msg['Subject'] = Header(subject, 'utf-8')
msg['From'] = Header(from, 'utf-8')
msg['To'] = Header(to, 'utf-8')
text = msg.as_string()
try:
server.send(from, emails, text)
print('Message Sent Succesfully')
except:
print('There Was An Error While Sending The Message')
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()
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())
