[python] How to send an email with Python?

This code works and sends me an email just fine:

import smtplib
#SERVER = "localhost"

FROM = '[email protected]'

TO = ["[email protected]"] # 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?

As far your code is concerned, there doesn't seem to be anything fundamentally wrong with it except that, it is unclear how you're actually calling that function. All I can think of is that when your server is not responding then you will get this SMTPServerDisconnected error. If you lookup the getreply() function in smtplib (excerpt below), you will get an idea.

def getreply(self):
    """Get a reply from the server.

    Returns a tuple consisting of:

      - server response code (e.g. '250', or such, if all goes well)
        Note: returns -1 if it can't read response code.

      - server response string corresponding to response code (multiline
        responses are converted to a single, multiline string).

    Raises SMTPServerDisconnected if end-of-file is reached.
    """

check an example at https://github.com/rreddy80/sendEmails/blob/master/sendEmailAttachments.py that also uses a function call to send an email, if that's what you're trying to do (DRY approach).

Thought I'd put in my two bits here since I have just figured out how this works.

It appears that you don't have the port specified on your SERVER connection settings, this effected me a little bit when I was trying to connect to my SMTP server that isn't using the default port: 25.

According to the smtplib.SMTP docs, your ehlo or helo request/response should automatically be taken care of, so you shouldn't have to worry about this (but might be something to confirm if all else fails).

Another thing to ask yourself is have you allowed SMTP connections on your SMTP server itself? For some sites like GMAIL and ZOHO you have to actually go in and activate the IMAP connections within the email account. Your mail server might not allow SMTP connections that don't come from 'localhost' perhaps? Something to look into.

The final thing is you might want to try and initiate the connection on TLS. Most servers now require this type of authentication.

You'll see I've jammed two TO fields into my email. The msg['TO'] and msg['FROM'] msg dictionary items allows the correct information to show up in the headers of the email itself, which one sees on the receiving end of the email in the To/From fields (you might even be able to add a Reply To field in here. The TO and FROM fields themselves are what the server requires. I know I've heard of some email servers rejecting emails if they don't have the proper email headers in place.

This is the code I've used, in a function, that works for me to email the content of a *.txt file using my local computer and a remote SMTP server (ZOHO as shown):

def emailResults(folder, filename):

    # body of the message
    doc = folder + filename + '.txt'
    with open(doc, 'r') as readText:
        msg = MIMEText(readText.read())

    # headers
    TO = '[email protected]'
    msg['To'] = TO
    FROM = '[email protected]'
    msg['From'] = FROM
    msg['Subject'] = 'email subject |' + filename

    # SMTP
    send = smtplib.SMTP('smtp.zoho.com', 587)
    send.starttls()
    send.login('[email protected]', 'password')
    send.sendmail(FROM, TO, msg.as_string())
    send.quit()

It's worth noting that the SMTP module supports the context manager so there is no need to manually call quit(), this will guarantee it is always called even if there is an exception.

    with smtplib.SMTP_SSL('smtp.gmail.com', 465) as server:
        server.ehlo()
        server.login(user, password)
        server.sendmail(from, to, body)

There is indentation problem. The code below will work:

import textwrap

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    import smtplib
    """this is some test documentation in the function"""
    message = textwrap.dedent("""\
        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()

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='[email protected]'):
    # 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=['[email protected]', '[email protected]'],
          subject='hello', message='Your analysis has done!')

below may only for Chinese user:

If you use 126/163, ????, you need to set"???????", like below:

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

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: [email protected]    To: [email protected]    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:

[email protected]    To: [email protected]    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: [email protected]
To: [email protected]
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).

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)

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.

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

Well, you want to have an answer that is up-to-date and modern.

Here is my answer:

When I need to mail in Python, I use the mailgun API wich get's a lot of the headaches with sending mails sorted out. They have a wonderfull app/api that allows you to send 5,000 free emails per month.

Sending an email would be like this:

def send_simple_message():
    return requests.post(
        "https://api.mailgun.net/v3/YOUR_DOMAIN_NAME/messages",
        auth=("api", "YOUR_API_KEY"),
        data={"from": "Excited User <mailgun@YOUR_DOMAIN_NAME>",
              "to": ["[email protected]", "YOU@YOUR_DOMAIN_NAME"],
              "subject": "Hello",
              "text": "Testing some Mailgun awesomness!"})

You can also track events and lots more, see the quickstart guide.

I hope you find this useful!