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?
只是为了补充答案,以便您的邮件传递系统可以扩展。
我建议有一个配置文件(可以是.json, .yml, .ini等),包含发件人的电子邮件配置,密码和收件人。
通过这种方式,您可以根据需要创建不同的可定制项目。
下面是一个包含3个文件,config, functions和main的小示例。纯文本邮件。
config_email.ini
[email_1]
sender = test@test.com
password = XXXXXXXXXXX
recipients= ["email_2@test.com", "email_2@test.com"]
[email_2]
sender = test_2@test.com
password = XXXXXXXXXXX
recipients= ["email_2@test.com", "email_2@test.com", "email_3@test.com"]
这些项将从main.py调用,它将返回它们各自的值。
函数functions_email.py文件:
import smtplib,configparser,json
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
def get_credentials(item):
parse = configparser.ConfigParser()
parse.read('config_email.ini')
sender = parse[item]['sender ']
password = parse[item]['password']
recipients= json.loads(parse[item]['recipients'])
return sender,password,recipients
def get_msg(sender,recipients,subject,mail_body):
msg = MIMEMultipart()
msg['Subject'] = subject
msg['From'] = sender
msg['To'] = ', '.join(recipients)
text = """\
"""+mail_body+""" """
part1 = MIMEText(text, "plain")
msg.attach(part1)
return msg
def send_email(msg,sender,password,recipients):
s = smtplib.SMTP('smtp.test.com')
s.login(sender,password)
s.sendmail(sender, recipients, msg.as_string())
s.quit()
文件main.py:
from functions_email import *
sender,password,recipients = get_credenciales('email_2')
subject= 'text to subject'
mail_body = 'body....................'
msg = get_msg(sender,recipients ,subject,mail_body)
send_email(msg,sender,password,recipients)
最好的问候!
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()
我想通过建议yagmail包来帮助你发送电子邮件(我是维护者,抱歉广告,但我觉得它真的能帮助!)
你的整个代码将是:
import yagmail
yag = yagmail.SMTP(FROM, 'pass')
yag.send(TO, SUBJECT, TEXT)
注意,我为所有参数提供了默认值,例如,如果你想发送给自己,你可以省略to,如果你不想要一个主题,你也可以省略它。
此外,我们的目标还在于使附加html代码或图像(以及其他文件)变得非常容易。
在你放置内容的地方,你可以这样做:
contents = ['Body text, and here is an embedded image:', 'http://somedomain/image.png',
'You can also find an audio file attached.', '/local/path/song.mp3']
哇,发送附件是多么简单啊!如果没有yagmail,这大概需要20行;)
此外,如果你设置了一次,你就永远不必再输入密码(并安全地保存密码)。在你的情况下,你可以这样做:
import yagmail
yagmail.SMTP().send(contents = contents)
这样更简洁!
我建议你看看github,或者直接用pip install yagmail安装它。
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")