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?
在缩进函数中的代码时(这是可以的)，还缩进了原始消息字符串的行。但是前导空白意味着标题行的折叠(连接)，如RFC 2822 - Internet Message Format的2.2.3和3.2.3节所述:
每个报头字段在逻辑上是由一行字符组成的
字段名、冒号和字段主体。为了方便
但是，为了处理每行998/78个字符的限制，
报头字段的字段主体部分可以分成多个
线表示;这叫做“折叠”。
在sendmail调用的函数形式中，所有行都以空白开始，因此是“展开的”(连接)，您正在尝试发送
From: monty@python.com To: jon@mycompany.com Subject: Hello! This message was sent with Python's smtplib.
与我们的想法不同，smtplib将不再理解To:和Subject:头文件，因为这些名称只在一行的开头被识别。相反，smtplib将假设一个非常长的发送者电子邮件地址:
monty@python.com To: jon@mycompany.com Subject: Hello! This message was sent with Python's smtplib.
这将不起作用，因此出现异常。
解决方案很简单:只保留原来的消息字符串。这可以通过一个函数来完成(正如Zeeshan建议的那样)，也可以直接在源代码中完成:
import smtplib
def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
"""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()
现在展开没有发生，你发送
From: monty@python.com
To: jon@mycompany.com
Subject: Hello!
This message was sent with Python's smtplib.
这就是您的旧代码所做的工作。
请注意，我还保留了标题和正文之间的空行，以适应RFC的第3.5节(这是必需的)，并根据Python风格指南PEP-0008(这是可选的)将include放在函数之外。
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!
