How to send an email with Python?

Question

This code works and sends me an email just fine:

import smtplib
#SERVER = "localhost"

FROM = '[e-mail]'

TO = ["[e-mail]"] # 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()

However if I try to wrap it in a function like this:

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()

and call it I get the following errors:

 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

Can anyone help me understand why?

Answer 1

While indenting your code in the function (which is ok), you did also indent the lines of the raw message string. But leading white space implies folding (concatenation) of the header lines, as described in sections 2.2.3 and 3.2.3 of RFC 2822 - Internet Message Format:

Each header field is logically a single line of characters comprising the field name, the colon, and the field body. For convenience however, and to deal with the 998/78 character limitations per line, the field body portion of a header field can be split into a multiple line representation; this is called "folding".

In the function form of your sendmail call, all lines are starting with white space and so are "unfolded" (concatenated) and you are trying to send

From: [e-mail]    To: [e-mail]    Subject: Hello!    This message was sent with Python's smtplib.

Other than our mind suggests, smtplib will not understand the To: and Subject: headers any longer, because these names are only recognized at the beginning of a line. Instead smtplib will assume a very long sender email address:

[e-mail]    To: [e-mail]    Subject: Hello!    This message was sent with Python's smtplib.

This won't work and so comes your Exception.

The solution is simple: Just preserve the message string as it was before. This can be done by a function (as Zeeshan suggested) or right away in the source code:

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()

Now the unfolding does not occur and you send

From: [e-mail]
To: [e-mail]
Subject: Hello!

This message was sent with Python's smtplib.

which is what works and what was done by your old code.

Note that I was also preserving the empty line between headers and body to accommodate section 3.5 of the RFC (which is required) and put the include outside the function according to the Python style guide PEP-0008 (which is optional).

Answer 2

I'd like to help you with sending emails by advising the yagmail package (I'm the maintainer, sorry for the advertising, but I feel it can really help!).

The whole code for you would be:

import yagmail
yag = yagmail.SMTP(FROM, 'pass')
yag.send(TO, SUBJECT, TEXT)

Note that I provide defaults for all arguments, for example if you want to send to yourself, you can omit TO, if you don't want a subject, you can omit it also.

Furthermore, the goal is also to make it really easy to attach html code or images (and other files).

Where you put contents you can do something like:

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']

Wow, how easy it is to send attachments! This would take like 20 lines without yagmail ;)

Also, if you set it up once, you'll never have to enter the password again (and have it safely stored). In your case you can do something like:

import yagmail
yagmail.SMTP().send(contents = contents)

which is much more concise!

I'd invite you to have a look at the github or install it directly with pip install yagmail.

Answer 3

I recommend that you use the standard packages email and smtplib together to send email. Please look at the following example (reproduced from the Python documentation). Notice that if you follow this approach, the "simple" task is indeed simple, and the more complex tasks (like attaching binary objects or sending plain/HTML multipart messages) are accomplished very rapidly.

# Import smtplib for the actual sending function
import smtplib

# Import the email modules we'll need
from email.mime.text import MIMEText

# Open a plain text file for reading.  For this example, assume that
# the text file contains only ASCII characters.
with open(textfile, 'rb') as fp:
    # Create a text/plain message
    msg = MIMEText(fp.read())

# me == the sender's email address
# you == the recipient's email address
msg['Subject'] = 'The contents of %s' % textfile
msg['From'] = me
msg['To'] = you

# Send the message via our own SMTP server, but don't include the
# envelope header.
s = smtplib.SMTP('localhost')
s.sendmail(me, [you], msg.as_string())
s.quit()

For sending email to multiple destinations, you can also follow the example in the Python documentation:

# Import smtplib for the actual sending function
import smtplib

# Here are the email package modules we'll need
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart

# Create the container (outer) email message.
msg = MIMEMultipart()
msg['Subject'] = 'Our family reunion'
# me == the sender's email address
# family = the list of all recipients' email addresses
msg['From'] = me
msg['To'] = ', '.join(family)
msg.preamble = 'Our family reunion'

# Assume we know that the image files are all in PNG format
for file in pngfiles:
    # Open the files in binary mode.  Let the MIMEImage class automatically
    # guess the specific image type.
    with open(file, 'rb') as fp:
        img = MIMEImage(fp.read())
    msg.attach(img)

# Send the email via our own SMTP server.
s = smtplib.SMTP('localhost')
s.sendmail(me, family, msg.as_string())
s.quit()

As you can see, the header To in the MIMEText object must be a string consisting of email addresses separated by commas. On the other hand, the second argument to the sendmail function must be a list of strings (each string is an email address).

So, if you have three email addresses: [e-mail], [e-mail], and [e-mail], you can do as follows (obvious sections omitted):

to = ["[e-mail]", "[e-mail]", "[e-mail]"]
msg['To'] = ",".join(to)
s.sendmail(me, to, msg.as_string())

the "","".join(to) part makes a single string out of the list, separated by commas.

From your questions I gather that you have not gone through the Python tutorial - it is a MUST if you want to get anywhere in Python - the documentation is mostly excellent for the standard library.

Answer 4

Make sure you have granted permission for both Sender and Receiver to send email and receive email from Unknown sources(External Sources) in Email Account.

import smtplib

#Ports 465 and 587 are intended for email client to email server communication - sending email
server = smtplib.SMTP('smtp.gmail.com', 587)

#starttls() is a way to take an existing insecure connection and upgrade it to a secure connection using SSL/TLS.
server.starttls()

#Next, log in to the server
server.login("#email", "#password")

msg = "Hello! This Message was sent by the help of Python"

#Send the mail
server.sendmail("#Sender", "#Reciever", msg)

Answer 5

It's probably putting tabs into your message. Print out message before you pass it to sendMail.

Answer 6

Here is an example on Python 3.x, much simpler than 2.x:

import smtplib
from email.message import EmailMessage
def send_mail(to_email, subject, message, server='smtp.example.cn',
              from_email='[e-mail]'):
    # import smtplib
    msg = EmailMessage()
    msg['Subject'] = subject
    msg['From'] = from_email
    msg['To'] = ', '.join(to_email)
    msg.set_content(message)
    print(msg)
    server = smtplib.SMTP(server)
    server.set_debuglevel(1)
    server.login(from_email, 'password')  # user & password
    server.send_message(msg)
    server.quit()
    print('successfully sent the mail.')

call this function:

send_mail(to_email=['[e-mail]', '[e-mail]'],
          subject='hello', message='Your analysis has done!')

below may only for Chinese user:

If you use 126/163, 网易邮箱, you need to set"客户端授权密码", like below:

ref: https://stackoverflow.com/a/41470149/2803344 https://docs.python.org/3/library/email.examples.html#email-examples