这段代码工作，并向我发送电子邮件就好:
import smtplib
#SERVER = "localhost"
FROM = 'monty@python.com'
TO = ["jon@mycompany.com"] # must be a list
SUBJECT = "Hello!"
TEXT = "This message was sent with Python's smtplib."
# Prepare actual message
message = """\
From: %s
To: %s
Subject: %s
%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)
# Send the mail
server = smtplib.SMTP('myserver')
server.sendmail(FROM, TO, message)
server.quit()
然而，如果我试图将它包装在这样一个函数中:
def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
import smtplib
"""this is some test documentation in the function"""
message = """\
From: %s
To: %s
Subject: %s
%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)
# Send the mail
server = smtplib.SMTP(SERVER)
server.sendmail(FROM, TO, message)
server.quit()
我得到以下错误:
Traceback (most recent call last):
File "C:/Python31/mailtest1.py", line 8, in <module>
sendmail.sendMail(sender,recipients,subject,body,server)
File "C:/Python31\sendmail.py", line 13, in sendMail
server.sendmail(FROM, TO, message)
File "C:\Python31\lib\smtplib.py", line 720, in sendmail
self.rset()
File "C:\Python31\lib\smtplib.py", line 444, in rset
return self.docmd("rset")
File "C:\Python31\lib\smtplib.py", line 368, in docmd
return self.getreply()
File "C:\Python31\lib\smtplib.py", line 345, in getreply
raise SMTPServerDisconnected("Connection unexpectedly closed")
smtplib.SMTPServerDisconnected: Connection unexpectedly closed
有人能告诉我为什么吗?
我想我应该把我的两个比特放在这里，因为我刚刚明白了它是如何工作的。
似乎你没有在你的服务器连接设置上指定端口，这影响了我一点，当我试图连接到我的SMTP服务器，没有使用默认端口:25。
根据smtplib。SMTP文档，您的ehlo或helo请求/响应应该自动处理，所以您不必担心这一点(但如果其他都失败了，可能需要确认)。
另一个问题是你是否允许在你的SMTP服务器上进行SMTP连接?对于像GMAIL和ZOHO这样的网站，你必须实际进入并激活电子邮件帐户中的IMAP连接。您的邮件服务器可能不允许SMTP连接不是来自'localhost'也许?一些值得调查的事情。
最后一件事是你可能想尝试在TLS上发起连接。现在大多数服务器都需要这种类型的身份验证。
您将看到我在电子邮件中插入了两个TO字段。msg['TO']和msg['FROM'] msg字典项允许正确的信息显示在电子邮件本身的标题中，这可以在电子邮件的接收端的TO / FROM字段中看到(你甚至可以在这里添加一个Reply TO字段)。TO和FROM字段本身就是服务器所需要的。我知道我听说过一些电子邮件服务器拒绝邮件，如果他们没有适当的电子邮件标题。
这是我使用的代码，在一个函数中，为我工作，使用我的本地计算机和远程SMTP服务器(ZOHO所示)发送*.txt文件的内容:
def emailResults(folder, filename):
# body of the message
doc = folder + filename + '.txt'
with open(doc, 'r') as readText:
msg = MIMEText(readText.read())
# headers
TO = 'to_user@domain.com'
msg['To'] = TO
FROM = 'from_user@domain.com'
msg['From'] = FROM
msg['Subject'] = 'email subject |' + filename
# SMTP
send = smtplib.SMTP('smtp.zoho.com', 587)
send.starttls()
send.login('from_user@domain.com', 'password')
send.sendmail(FROM, TO, msg.as_string())
send.quit()
在摆弄了很多例子之后，例如这里
这对我来说很管用:
import smtplib
from email.mime.text import MIMEText
# SMTP sendmail server mail relay
host = 'mail.server.com'
port = 587 # starttls not SSL 465 e.g gmail, port 25 blocked by most ISPs & AWS
sender_email = 'name@server.com'
recipient_email = 'name@domain.com'
password = 'YourSMTPServerAuthenticationPass'
subject = "Server - "
body = "Message from server"
def sendemail(host, port, sender_email, recipient_email, password, subject, body):
try:
p1 = f'<p><HR><BR>{recipient_email}<BR>'
p2 = f'<h2><font color="green">{subject}</font></h2>'
p3 = f'<p>{body}'
p4 = f'<p>Kind Regards,<BR><BR>{sender_email}<BR><HR>'
message = MIMEText((p1+p2+p3+p4), 'html')
# servers may not accept non RFC 5321 / RFC 5322 / compliant TXT & HTML typos
message['From'] = f'Sender Name <{sender_email}>'
message['To'] = f'Receiver Name <{recipient_email}>'
message['Cc'] = f'Receiver2 Name <>'
message['Subject'] = f'{subject}'
msg = message.as_string()
server = smtplib.SMTP(host, port)
print("Connection Status: Connected")
server.set_debuglevel(1)
server.ehlo()
server.starttls()
server.ehlo()
server.login(sender_email, password)
print("Connection Status: Logged in")
server.sendmail(sender_email, recipient_email, msg)
print("Status: Email as HTML successfully sent")
except Exception as e:
print(e)
print("Error: unable to send email")
# Run
sendemail(host, port, sender_email, recipient_email, password, subject, body)
print("Status: Exit")
