Python 实用宝典

Question 1

I need for testing purposes to populate few hundred email boxes with various messages, and was going to use smtplib for that. But among other things I need to be able to send messages not only TO specific mailboxes, but CC and BCC them as well. It does not look like smtplib supports CC-ing and BCC-ing while sending emails.

Looking for suggestions how to do CC or BCC sending messages from the python script.

(And — no, I’m not creating a script to spam anyone outside of my testing environment.)

Question 2

Email headers don’t matter to the smtp server. Just add the CC and BCC recipients to the toaddrs when you send your email. For CC, add them to the CC header.

toaddr = 'buffy@sunnydale.k12.ca.us'
cc = ['alexander@sunydale.k12.ca.us','willow@sunnydale.k12.ca.us']
bcc = ['chairman@slayerscouncil.uk']
fromaddr = 'giles@sunnydale.k12.ca.us'
message_subject = "disturbance in sector 7"
message_text = "Three are dead in an attack in the sewers below sector 7."
message = "From: %s\r\n" % fromaddr
        + "To: %s\r\n" % toaddr
        + "CC: %s\r\n" % ",".join(cc)
        + "Subject: %s\r\n" % message_subject
        + "\r\n" 
        + message_text
toaddrs = [toaddr] + cc + bcc
server = smtplib.SMTP('smtp.sunnydale.k12.ca.us')
server.set_debuglevel(1)
server.sendmail(fromaddr, toaddrs, message)
server.quit()

Question 3

Key thing is to add the recipients as a list of email ids in your sendmail call.

import smtplib
from email.mime.multipart import MIMEMultipart

me = "user63503@gmail.com"
to = "someone@gmail.com"
cc = "anotherperson@gmail.com,someone@yahoo.com"
bcc = "bccperson1@gmail.com,bccperson2@yahoo.com"

rcpt = cc.split(",") + bcc.split(",") + [to]
msg = MIMEMultipart('alternative')
msg['Subject'] = "my subject"
msg['To'] = to
msg['Cc'] = cc
msg.attach(my_msg_body)
server = smtplib.SMTP("localhost") # or your smtp server
server.sendmail(me, rcpt, msg.as_string())
server.quit()

Question 4

Don’t add the bcc header.

See this: http://mail.python.org/pipermail/email-sig/2004-September/000151.html

And this: “””Notice that the second argument to sendmail(), the recipients, is passed as a list. You can include any number of addresses in the list to have the message delivered to each of them in turn. Since the envelope information is separate from the message headers, you can even BCC someone by including them in the method argument but not in the message header.””” from http://pymotw.com/2/smtplib

toaddr = 'buffy@sunnydale.k12.ca.us'
cc = ['alexander@sunydale.k12.ca.us','willow@sunnydale.k12.ca.us']
bcc = ['chairman@slayerscouncil.uk']
fromaddr = 'giles@sunnydale.k12.ca.us'
message_subject = "disturbance in sector 7"
message_text = "Three are dead in an attack in the sewers below sector 7."
message = "From: %s\r\n" % fromaddr
    + "To: %s\r\n" % toaddr
    + "CC: %s\r\n" % ",".join(cc)
    # don't add this, otherwise "to and cc" receivers will know who are the bcc receivers
    # + "BCC: %s\r\n" % ",".join(bcc)
    + "Subject: %s\r\n" % message_subject
    + "\r\n" 
    + message_text
toaddrs = [toaddr] + cc + bcc
server = smtplib.SMTP('smtp.sunnydale.k12.ca.us')
server.set_debuglevel(1)
server.sendmail(fromaddr, toaddrs, message)
server.quit()

Question 5

As of Python 3.2, released Nov 2011, the smtplib has a new function send_message instead of just sendmail, which makes dealing with To/CC/BCC easier. Pulling from the Python official email examples, with some slight modifications, we get:

# Import smtplib for the actual sending function
import smtplib

# Import the email modules we'll need
from email.message import EmailMessage

# Open the plain text file whose name is in textfile for reading.
with open(textfile) as fp:
    # Create a text/plain message
    msg = EmailMessage()
    msg.set_content(fp.read())

# me == the sender's email address
# you == the recipient's email address
# them == the cc's email address
# they == the bcc's email address
msg['Subject'] = 'The contents of %s' % textfile
msg['From'] = me
msg['To'] = you
msg['Cc'] = them
msg['Bcc'] = they


# Send the message via our own SMTP server.
s = smtplib.SMTP('localhost')
s.send_message(msg)
s.quit()

Using the headers work fine, because send_message respects BCC as outlined in the documentation:

send_message does not transmit any Bcc or Resent-Bcc headers that may appear in msg

With sendmail it was common to add the CC headers to the message, doing something such as:

msg['Bcc'] = blind.email@adrress.com

Or

msg = "From: from.email@address.com" +
      "To: to.email@adress.com" +
      "BCC: hidden.email@address.com" +
      "Subject: You've got mail!" +
      "This is the message body"

The problem is, the sendmail function treats all those headers the same, meaning they’ll get sent (visibly) to all To: and BCC: users, defeating the purposes of BCC. The solution, as shown in many of the other answers here, was to not include BCC in the headers, and instead only in the list of emails passed to sendmail.

The caveat is that send_message requires a Message object, meaning you’ll need to import a class from email.message instead of merely passing strings into sendmail.

Question 6

The distinction between TO, CC and BCC occurs only in the text headers. At the SMTP level, everybody is a recipient.

TO – There is a TO: header with this recipient’s address

CC – There is a CC: header with this recipient’s address

BCC – This recipient isn’t mentioned in the headers at all, but is still a recipient.

If you have

TO: abc@company.com
CC: xyz@company.com
BCC: boss@company.com

You have three recipients. The headers in the email body will include only the TO: and CC:

Question 7

You can try MIMEText

msg = MIMEText('text')
msg['to'] = 
msg['cc'] =

then send msg.as_string()

https://docs.python.org/3.6/library/email.examples.html

Question 8

It did not worked for me until i created:

#created cc string
cc = ""someone@domain.com;
#added cc to header
msg['Cc'] = cc

and than added cc in recipient [list] like:

s.sendmail(me, [you,cc], msg.as_string())

Question 9

None of the above things worked for me as I had multiple recipients both in ‘to’ and ‘cc’. So I tried like below:

recipients = ['abc@gmail.com', 'xyz@gmail.com']
cc_recipients = ['lmn@gmail.com', 'pqr@gmail.com']
MESSAGE['To'] = ", ".join(recipients)
MESSAGE['Cc'] = ", ".join(cc_recipients)

and extend the ‘recipients’ with ‘cc_recipients’ and send mail in trivial way

recipients.extend(cc_recipients)
server.sendmail(FROM,recipients,MESSAGE.as_string())

Question 10

This code works and sends me an email just fine:

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()

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?

Question 11

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: person1@example.com, person2@example.com, and person3@example.com, you can do as follows (obvious sections omitted):

to = ["person1@example.com", "person2@example.com", "person3@example.com"]
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.

Question 12

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": ["bar@example.com", "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!

Question 13

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.

Question 14

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()

Question 15

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='xx@example.com'):
    # 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=['12345@qq.com', '12345@126.com'],
          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

Question 16

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: monty@python.com    To: jon@mycompany.com    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:

monty@python.com    To: jon@mycompany.com    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: monty@python.com
To: jon@mycompany.com
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).

Question 17

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

Question 18

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)

Question 19

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 = '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()

Question 20

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)

Question 21

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).

Question 22

In my settings.py, I have the following:

EMAIL_BACKEND = 'django.core.mail.backends.smtp.EmailBackend'

# Host for sending e-mail.
EMAIL_HOST = 'localhost'

# Port for sending e-mail.
EMAIL_PORT = 1025

# Optional SMTP authentication information for EMAIL_HOST.
EMAIL_HOST_USER = ''
EMAIL_HOST_PASSWORD = ''
EMAIL_USE_TLS = False

My email code:

from django.core.mail import EmailMessage
email = EmailMessage('Hello', 'World', to=['user@gmail.com'])
email.send()

Of course, if I setup a debugging server via python -m smtpd -n -c DebuggingServer localhost:1025, I can see the email in my terminal.

However, how do I actually send the email not to the debugging server but to user@gmail.com?

After reading your answers, let me get something straight:

Can’t you use localhost(simple ubuntu pc) to send e-mails?
I thought in django 1.3 send_mail() is somewhat deprecated and EmailMessage.send() is used instead?

Question 23

Send the email to a real SMTP server. If you don’t want to set up your own then you can find companies that will run one for you, such as Google themselves.

Question 24

I use Gmail as my SMTP server for Django. Much easier than dealing with postfix or whatever other server. I’m not in the business of managing email servers.

In settings.py:

EMAIL_USE_TLS = True
EMAIL_HOST = 'smtp.gmail.com'
EMAIL_PORT = 587
EMAIL_HOST_USER = 'me@gmail.com'
EMAIL_HOST_PASSWORD = 'password'

NOTE: In 2016 Gmail is not allowing this anymore by default. You can either use an external service like Sendgrid, or you can follow this tutorial from Google to reduce security but allow this option: https://support.google.com/accounts/answer/6010255

Question 25

Create a project: django-admin.py startproject gmail

Edit settings.py with code below:

EMAIL_BACKEND = 'django.core.mail.backends.smtp.EmailBackend'
EMAIL_USE_TLS = True
EMAIL_HOST = 'smtp.gmail.com'
EMAIL_HOST_USER = 'youremail@gmail.com'
EMAIL_HOST_PASSWORD = 'email_password'
EMAIL_PORT = 587

Run interactive mode: python manage.py shell

Import the EmailMessage module:

from django.core.mail import EmailMessage

Send the email:

email = EmailMessage('Subject', 'Body', to=['your@email.com'])
email.send()

For more informations, check send_mail and EmailMessage features in documents.

UPDATE for Gmail

Also if you have problems sending email via gmail remember to check this guides from google.

In your Google account settings, go to Security > Account permissions > Access for less secure apps and enable this option.

Also create an App specific password for your gmail after you’ve turned on 2-step-verification for it.

Then you should use app specific password in settings. So change the following line:

    EMAIL_HOST_PASSWORD = 'your_email_app_specific_password'

Also if you’re interested to send HTML email, check this out.

Question 26

My site is hosted on Godaddy and I have a private email registered on the same. These are the settings which worked for me:

In settings.py:

EMAIL_HOST = 'mail.domain.com'
EMAIL_HOST_USER = 'abc@domain.com'
EMAIL_HOST_PASSWORD = 'abcdef'
DEFAULT_FROM_EMAIL = 'abc@domain.com'
SERVER_EMAIL = 'abc@domain.com'
EMAIL_PORT = 25
EMAIL_USE_TLS = False

In shell:

from django.core.mail import EmailMessage
email = EmailMessage('Subject', 'Body', to=['def@domain.com'])
email.send()

Then I got “1” as the O/P i.e. Success. And I received the mail too. :)

What is the meaning of domain.com?

Question 27

For Django version 1.7, if above solutions dont work then try the following

in settings.py add

#For email
EMAIL_BACKEND = 'django.core.mail.backends.smtp.EmailBackend'

EMAIL_USE_TLS = True

EMAIL_HOST = 'smtp.gmail.com'

EMAIL_HOST_USER = 'sender@gmail.com'

#Must generate specific password for your app in [gmail settings][1]
EMAIL_HOST_PASSWORD = 'app_specific_password'

EMAIL_PORT = 587

#This did the trick
DEFAULT_FROM_EMAIL = EMAIL_HOST_USER

The last line did the trick for django 1.7

Question 28

You need to use smtp as backend in settings.py

EMAIL_BACKEND = 'django.core.mail.backends.smtp.EmailBackend'

If you use backend as console, you will receive output in console

EMAIL_BACKEND = 'django.core.mail.backends.console.EmailBackend'

And also below settings in addition

EMAIL_USE_TLS = True
EMAIL_HOST = 'smtp.gmail.com'
EMAIL_PORT = 587
EMAIL_HOST_USER = 'urusername@gmail.com'
EMAIL_HOST_PASSWORD = 'password'

If you are using gmail for this, setup 2-step verification and Application specific password and copy and paste that password in above EMAIL_HOST_PASSWORD value.

Question 29

I found using SendGrid to be the easiest way to set up sending email with Django. Here’s how it works:

Create a SendGrid account (and verify your email)
Add the following to your settings.py: EMAIL_HOST = 'smtp.sendgrid.net' EMAIL_HOST_USER = '<your sendgrid username>' EMAIL_HOST_PASSWORD = '<your sendgrid password>' EMAIL_PORT = 587 EMAIL_USE_TLS = True

And you’re all set!

To send email:

from django.core.mail import send_mail
send_mail('<Your subject>', '<Your message>', 'from@example.com', ['to@example.com'])

If you want Django to email you whenever there’s a 500 internal server error, add the following to your settings.py:

DEFAULT_FROM_EMAIL = 'your.email@example.com'
ADMINS = [('<Your name>', 'your.email@example.com')]

Sending email with SendGrid is free up to 12k emails per month.

Question 30

I had actually done this from Django a while back. Open up a legitimate GMail account & enter the credentials here. Here’s my code –

from email import Encoders
from email.MIMEBase import MIMEBase
from email.MIMEText import MIMEText
from email.MIMEMultipart import MIMEMultipart

def sendmail(to, subject, text, attach=[], mtype='html'):
    ok = True
    gmail_user = settings.EMAIL_HOST_USER
    gmail_pwd  = settings.EMAIL_HOST_PASSWORD

    msg = MIMEMultipart('alternative')

    msg['From']    = gmail_user
    msg['To']      = to
    msg['Cc']      = 'you@gmail.com'
    msg['Subject'] = subject

    msg.attach(MIMEText(text, mtype))

    for a in attach:
        part = MIMEBase('application', 'octet-stream')
        part.set_payload(open(attach, 'rb').read())
        Encoders.encode_base64(part)
        part.add_header('Content-Disposition','attachment; filename="%s"' % os.path.basename(a))
        msg.attach(part)

    try:
        mailServer = smtplib.SMTP("smtp.gmail.com", 687)
        mailServer.ehlo()
        mailServer.starttls()
        mailServer.ehlo()
        mailServer.login(gmail_user, gmail_pwd)
        mailServer.sendmail(gmail_user, [to,msg['Cc']], msg.as_string())
        mailServer.close()
    except:
        ok = False
    return ok

Question 31

Late, but:

In addition to the DEFAULT_FROM_EMAIL fix others have mentioned, and allowing less-secure apps to access the account, I had to navigate to https://accounts.google.com/DisplayUnlockCaptcha while signed in as the account in question to get Django to finally authenticate.

I went to that URL through a SSH tunnel to the web server to make sure the IP address was the same; I’m not totally sure if that’s necessary but it can’t hurt. You can do that like so: ssh -D 8080 -fN <username>@<host>, then set your web browser to use localhost:8080 as a SOCKS proxy.

Question 32

You could use “Test Mail Server Tool” to test email sending on your machine or localhost. Google and Download “Test Mail Server Tool” and set it up.

Then in your settings.py:

EMAIL_BACKEND= 'django.core.mail.backends.smtp.EmailBackend'
EMAIL_HOST = 'localhost'
EMAIL_PORT = 25

From shell:

from django.core.mail import send_mail
send_mail('subject','message','sender email',['receipient email'],    fail_silently=False)

Question 33

For SendGrid – Django Specifically:

SendGrid Django Docs here

Set these variables in

settings.py

EMAIL_HOST = 'smtp.sendgrid.net'
EMAIL_HOST_USER = 'sendgrid_username'
EMAIL_HOST_PASSWORD = 'sendgrid_password'
EMAIL_PORT = 587
EMAIL_USE_TLS = True

in views.py

from django.core.mail import send_mail
send_mail('Subject here', 'Here is the message.', 'from@example.com', ['to@example.com'], fail_silently=False)

Question 34

After much searching I couldn’t find out how to use smtplib.sendmail to send to multiple recipients. The problem was every time the mail would be sent the mail headers would appear to contain multiple addresses, but in fact only the first recipient would receive the email.

The problem seems to be that the email.Message module expects something different than the smtplib.sendmail() function.

In short, to send to multiple recipients you should set the header to be a string of comma delimited email addresses. The sendmail() parameter to_addrs however should be a list of email addresses.

from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
import smtplib

msg = MIMEMultipart()
msg["Subject"] = "Example"
msg["From"] = "me@example.com"
msg["To"] = "malcom@example.com,reynolds@example.com,firefly@example.com"
msg["Cc"] = "serenity@example.com,inara@example.com"
body = MIMEText("example email body")
msg.attach(body)
smtp = smtplib.SMTP("mailhost.example.com", 25)
smtp.sendmail(msg["From"], msg["To"].split(",") + msg["Cc"].split(","), msg.as_string())
smtp.quit()

Question 35

This really works, I spent a lot of time trying multiple variants.

import smtplib
from email.mime.text import MIMEText

s = smtplib.SMTP('smtp.uk.xensource.com')
s.set_debuglevel(1)
msg = MIMEText("""body""")
sender = 'me@example.com'
recipients = ['john.doe@example.com', 'john.smith@example.co.uk']
msg['Subject'] = "subject line"
msg['From'] = sender
msg['To'] = ", ".join(recipients)
s.sendmail(sender, recipients, msg.as_string())

Question 36

The msg['To'] needs to be a string:

msg['To'] = "a@b.com, b@b.com, c@b.com"

While the recipients in sendmail(sender, recipients, message) needs to be a list:

sendmail("a@a.com", ["a@b.com", "b@b.com", "c@b.com"], "Howdy")

Question 37

You need to understand the difference between the visible address of an email, and the delivery.

msg["To"] is essentially what is printed on the letter. It doesn’t actually have any effect. Except that your email client, just like the regular post officer, will assume that this is who you want to send the email to.

The actual delivery however can work quite different. So you can drop the email (or a copy) into the post box of someone completely different.

There are various reasons for this. For example forwarding. The To: header field doesn’t change on forwarding, however the email is dropped into a different mailbox.

The smtp.sendmail command now takes care of the actual delivery. email.Message is the contents of the letter only, not the delivery.

In low-level SMTP, you need to give the receipients one-by-one, which is why a list of adresses (not including names!) is the sensible API.

For the header, it can also contain for example the name, e.g. To: First Last <email@addr.tld>, Other User <other@mail.tld>. Your code example therefore is not recommended, as it will fail delivering this mail, since just by splitting it on , you still not not have the valid adresses!

Question 38

It works for me.

import smtplib
from email.mime.text import MIMEText

s = smtplib.SMTP('smtp.uk.xensource.com')
s.set_debuglevel(1)
msg = MIMEText("""body""")
sender = 'me@example.com'
recipients = 'john.doe@example.com,john.smith@example.co.uk'
msg['Subject'] = "subject line"
msg['From'] = sender
msg['To'] = recipients
s.sendmail(sender, recipients.split(','), msg.as_string())

Question 39

I tried the below and it worked like a charm :)

rec_list =  ['first@example.com', 'second@example.com']
rec =  ', '.join(rec_list)

msg['To'] = rec

send_out = smtplib.SMTP('localhost')
send_out.sendmail(me, rec_list, msg.as_string())

Question 40

So actually the problem is that SMTP.sendmail and email.MIMEText need two different things.

email.MIMEText sets up the “To:” header for the body of the e-mail. It is ONLY used for displaying a result to the human being at the other end, and like all e-mail headers, must be a single string. (Note that it does not actually have to have anything to do with the people who actually receive the message.)

SMTP.sendmail, on the other hand, sets up the “envelope” of the message for the SMTP protocol. It needs a Python list of strings, each of which has a single address.

So, what you need to do is COMBINE the two replies you received. Set msg[‘To’] to a single string, but pass the raw list to sendmail:

emails = ['a.com','b.com', 'c.com']
msg['To'] = ', '.join( emails ) 
....
s.sendmail( msg['From'], emails, msg.as_string())

Question 41

I came up with this importable module function. It uses the gmail email server in this example. Its split into header and message so you can clearly see whats going on:

import smtplib

def send_alert(subject=""):

    to = ['email@one.com', 'email2@another_email.com', 'a3rd@email.com']
    gmail_user = 'me@gmail.com'
    gmail_pwd = 'my_pass'
    smtpserver = smtplib.SMTP("smtp.gmail.com", 587)
    smtpserver.ehlo()
    smtpserver.starttls()
    smtpserver.ehlo
    smtpserver.login(gmail_user, gmail_pwd)
    header = 'To:' + ", ".join(to) + '\n' + 'From: ' + gmail_user + '\n' + 'Subject: ' + subject + '\n'
    msg = header + '\n' + subject + '\n\n'
    smtpserver.sendmail(gmail_user, to, msg)
    smtpserver.close()

Question 42

I figured this out a few months back and blogged about it. The summary is:

If you want to use smtplib to send email to multiple recipients, use email.Message.add_header('To', eachRecipientAsString) to add them, and then when you invoke the sendmail method, use email.Message.get_all('To') send the message to all of them. Ditto for Cc and Bcc recipients.

Question 43

The solution below worked for me. It successfully sends an email to multiple recipients, including “CC” and “BCC.”

toaddr = ['mailid_1','mailid_2']
cc = ['mailid_3','mailid_4']
bcc = ['mailid_5','mailid_6']
subject = 'Email from Python Code'
fromaddr = 'sender_mailid'
message = "\n  !! Hello... !!"

msg['From'] = fromaddr
msg['To'] = ', '.join(toaddr)
msg['Cc'] = ', '.join(cc)
msg['Bcc'] = ', '.join(bcc)
msg['Subject'] = subject

s.sendmail(fromaddr, (toaddr+cc+bcc) , message)

Question 44

Well, the method in this asnwer method did not work for me. I don’t know, maybe this is a Python3 (I am using the 3.4 version) or gmail related issue, but after some tries, the solution that worked for me, was the line

s.send_message(msg)

instead of

s.sendmail(sender, recipients, msg.as_string())

Question 45

I use python 3.6 and the following code works for me

email_send = 'xxxxx@xxx.xxx,xxxx@xxx.xxx'
server.sendmail(email_user,email_send.split(','),text)

Question 46

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def sender(recipients): 

    body = 'Your email content here'
    msg = MIMEMultipart()

    msg['Subject'] = 'Email Subject'
    msg['From'] = 'your.email@gmail.com'
    msg['To'] = (', ').join(recipients.split(','))

    msg.attach(MIMEText(body,'plain'))

    server = smtplib.SMTP('smtp.gmail.com', 587)
    server.starttls()
    server.login('your.email@gmail.com', 'yourpassword')
    server.send_message(msg)
    server.quit()

if __name__ == '__main__':
    sender('email_1@domain.com,email_2@domain.com')

It only worked for me with send_message function and using the join function in the list whith recipients, python 3.6.

Question 47

you can try this when you write the recpient emails on a text file

from email.mime.text import MIMEText
from email.header import Header
import smtplib

f =  open('emails.txt', 'r').readlines()
for n in f:
     emails = n.rstrip()
server = smtplib.SMTP('smtp.uk.xensource.com')
server.ehlo()
server.starttls()
body = "Test Email"
subject = "Test"
from = "me@example.com"
to = emails
msg = MIMEText(body,'plain','utf-8')
msg['Subject'] = Header(subject, 'utf-8')
msg['From'] =  Header(from, 'utf-8')
msg['To'] = Header(to, 'utf-8')
text = msg.as_string()
try:
   server.send(from, emails, text)
   print('Message Sent Succesfully')
except:
   print('There Was An Error While Sending The Message')

Question 48

I am having problems understanding how to email an attachment using Python. I have successfully emailed simple messages with the smtplib. Could someone please explain how to send an attachment in an email. I know there are other posts online but as a Python beginner I find them hard to understand.

Question 49

Here’s another:

import smtplib
from os.path import basename
from email.mime.application import MIMEApplication
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.utils import COMMASPACE, formatdate


def send_mail(send_from, send_to, subject, text, files=None,
              server="127.0.0.1"):
    assert isinstance(send_to, list)

    msg = MIMEMultipart()
    msg['From'] = send_from
    msg['To'] = COMMASPACE.join(send_to)
    msg['Date'] = formatdate(localtime=True)
    msg['Subject'] = subject

    msg.attach(MIMEText(text))

    for f in files or []:
        with open(f, "rb") as fil:
            part = MIMEApplication(
                fil.read(),
                Name=basename(f)
            )
        # After the file is closed
        part['Content-Disposition'] = 'attachment; filename="%s"' % basename(f)
        msg.attach(part)


    smtp = smtplib.SMTP(server)
    smtp.sendmail(send_from, send_to, msg.as_string())
    smtp.close()

It’s much the same as the first example… But it should be easier to drop in.

Question 50

Here is the modified version from Oli for python 3

import smtplib
from pathlib import Path
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
from email.mime.text import MIMEText
from email.utils import COMMASPACE, formatdate
from email import encoders


def send_mail(send_from, send_to, subject, message, files=[],
              server="localhost", port=587, username='', password='',
              use_tls=True):
    """Compose and send email with provided info and attachments.

    Args:
        send_from (str): from name
        send_to (list[str]): to name(s)
        subject (str): message title
        message (str): message body
        files (list[str]): list of file paths to be attached to email
        server (str): mail server host name
        port (int): port number
        username (str): server auth username
        password (str): server auth password
        use_tls (bool): use TLS mode
    """
    msg = MIMEMultipart()
    msg['From'] = send_from
    msg['To'] = COMMASPACE.join(send_to)
    msg['Date'] = formatdate(localtime=True)
    msg['Subject'] = subject

    msg.attach(MIMEText(message))

    for path in files:
        part = MIMEBase('application', "octet-stream")
        with open(path, 'rb') as file:
            part.set_payload(file.read())
        encoders.encode_base64(part)
        part.add_header('Content-Disposition',
                        'attachment; filename="{}"'.format(Path(path).name))
        msg.attach(part)

    smtp = smtplib.SMTP(server, port)
    if use_tls:
        smtp.starttls()
    smtp.login(username, password)
    smtp.sendmail(send_from, send_to, msg.as_string())
    smtp.quit()

Question 51

This is the code I ended up using:

import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEBase import MIMEBase
from email import Encoders


SUBJECT = "Email Data"

msg = MIMEMultipart()
msg['Subject'] = SUBJECT 
msg['From'] = self.EMAIL_FROM
msg['To'] = ', '.join(self.EMAIL_TO)

part = MIMEBase('application', "octet-stream")
part.set_payload(open("text.txt", "rb").read())
Encoders.encode_base64(part)
    
part.add_header('Content-Disposition', 'attachment; filename="text.txt"')

msg.attach(part)

server = smtplib.SMTP(self.EMAIL_SERVER)
server.sendmail(self.EMAIL_FROM, self.EMAIL_TO, msg.as_string())

Code is much the same as Oli’s post.

Code based from Binary file email attachment problem post.

Question 52

from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
from email.MIMEImage import MIMEImage
import smtplib

msg = MIMEMultipart()
msg.attach(MIMEText(file("text.txt").read()))
msg.attach(MIMEImage(file("image.png").read()))

# to send
mailer = smtplib.SMTP()
mailer.connect()
mailer.sendmail(from_, to, msg.as_string())
mailer.close()

Adapted from here.

Question 53

Another way with python 3 (If someone is searching):

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders

fromaddr = "sender mail address"
toaddr = "receiver mail address"

msg = MIMEMultipart()

msg['From'] = fromaddr
msg['To'] = toaddr
msg['Subject'] = "SUBJECT OF THE EMAIL"

body = "TEXT YOU WANT TO SEND"

msg.attach(MIMEText(body, 'plain'))

filename = "fileName"
attachment = open("path of file", "rb")

part = MIMEBase('application', 'octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition', "attachment; filename= %s" % filename)

msg.attach(part)

server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login(fromaddr, "sender mail password")
text = msg.as_string()
server.sendmail(fromaddr, toaddr, text)
server.quit()

Make sure to allow “less secure apps” on your Gmail account

Question 54

Gmail version, working with Python 3.6 (note that you will need to change your Gmail settings to be able to send email via smtp from it:

import smtplib
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.application import MIMEApplication
from os.path import basename


def send_mail(send_from: str, subject: str, text: str, 
send_to: list, files= None):

    send_to= default_address if not send_to else send_to

    msg = MIMEMultipart()
    msg['From'] = send_from
    msg['To'] = ', '.join(send_to)  
    msg['Subject'] = subject

    msg.attach(MIMEText(text))

    for f in files or []:
        with open(f, "rb") as fil: 
            ext = f.split('.')[-1:]
            attachedfile = MIMEApplication(fil.read(), _subtype = ext)
            attachedfile.add_header(
                'content-disposition', 'attachment', filename=basename(f) )
        msg.attach(attachedfile)


    smtp = smtplib.SMTP(host="smtp.gmail.com", port= 587) 
    smtp.starttls()
    smtp.login(username,password)
    smtp.sendmail(send_from, send_to, msg.as_string())
    smtp.close()

Usage:

username = 'my-address@gmail.com'
password = 'top-secret'
default_address = ['my-address2@gmail.com'] 

send_mail(send_from= username,
subject="test",
text="text",
send_to= None,
files= # pass a list with the full filepaths here...
)

To use with any other email provider, just change the smtp configurations.

Question 55

The simplest code I could get to is:

#for attachment email
from django.core.mail import EmailMessage

    def attachment_email(request):
            email = EmailMessage(
            'Hello', #subject
            'Body goes here', #body
            'MyEmail@MyEmail.com', #from
            ['SendTo@SendTo.com'], #to
            ['bcc@example.com'], #bcc
            reply_to=['other@example.com'],
            headers={'Message-ID': 'foo'},
            )

            email.attach_file('/my/path/file')
            email.send()

It was based on the official Django documentation

Question 56

Other answers are excellent, though I still wanted to share a different approach in case someone is looking for alternatives.

Main difference here is that using this approach you can use HTML/CSS to format your message, so you can get creative and give some styling to your email. Though you aren’t enforced to use HTML, you can also still use only plain text.

Notice that this function accepts sending the email to multiple recipients and also allows to attach multiple files.

I’ve only tried this on Python 2, but I think it should work fine on 3 as well:

import os.path
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.application import MIMEApplication

def send_email(subject, message, from_email, to_email=[], attachment=[]):
    """
    :param subject: email subject
    :param message: Body content of the email (string), can be HTML/CSS or plain text
    :param from_email: Email address from where the email is sent
    :param to_email: List of email recipients, example: ["a@a.com", "b@b.com"]
    :param attachment: List of attachments, exmaple: ["file1.txt", "file2.txt"]
    """
    msg = MIMEMultipart()
    msg['Subject'] = subject
    msg['From'] = from_email
    msg['To'] = ", ".join(to_email)
    msg.attach(MIMEText(message, 'html'))

    for f in attachment:
        with open(f, 'rb') as a_file:
            basename = os.path.basename(f)
            part = MIMEApplication(a_file.read(), Name=basename)

        part['Content-Disposition'] = 'attachment; filename="%s"' % basename
        msg.attach(part)

    email = smtplib.SMTP('your-smtp-host-name.com')
    email.sendmail(from_email, to_email, msg.as_string())

I hope this helps! :-)

Question 57

from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
import smtplib
import mimetypes
import email.mime.application

smtp_ssl_host = 'smtp.gmail.com'  # smtp.mail.yahoo.com
smtp_ssl_port = 465
s = smtplib.SMTP_SSL(smtp_ssl_host, smtp_ssl_port)
s.login(email_user, email_pass)


msg = MIMEMultipart()
msg['Subject'] = 'I have a picture'
msg['From'] = email_user
msg['To'] = email_user

txt = MIMEText('I just bought a new camera.')
msg.attach(txt)

filename = 'introduction-to-algorithms-3rd-edition-sep-2010.pdf' #path to file
fo=open(filename,'rb')
attach = email.mime.application.MIMEApplication(fo.read(),_subtype="pdf")
fo.close()
attach.add_header('Content-Disposition','attachment',filename=filename)
msg.attach(attach)
s.send_message(msg)
s.quit()

For explanation, you can use this link it explains properly https://medium.com/@sdoshi579/to-send-an-email-along-with-attachment-using-smtp-7852e77623

Question 58

from email.mime.multipart import MIMEMultipart
from email.mime.image import MIMEImage
from email.mime.text import MIMEText
import smtplib

msg = MIMEMultipart()

password = "password"
msg['From'] = "from_address"
msg['To'] = "to_address"
msg['Subject'] = "Attached Photo"
msg.attach(MIMEImage(file("abc.jpg").read()))
file = "file path"
fp = open(file, 'rb')
img = MIMEImage(fp.read())
fp.close()
msg.attach(img)
server = smtplib.SMTP('smtp.gmail.com: 587')
server.starttls()
server.login(msg['From'], password)
server.sendmail(msg['From'], msg['To'], msg.as_string())
server.quit()

Question 59

Below is combination of what I’ve found from SoccerPlayer’s post Here and the following link that made it easier for me to attach an xlsx file. Found Here

file = 'File.xlsx'
username=''
password=''
send_from = ''
send_to = 'recipient1 , recipient2'
Cc = 'recipient'
msg = MIMEMultipart()
msg['From'] = send_from
msg['To'] = send_to
msg['Cc'] = Cc
msg['Date'] = formatdate(localtime = True)
msg['Subject'] = ''
server = smtplib.SMTP('smtp.gmail.com')
port = '587'
fp = open(file, 'rb')
part = MIMEBase('application','vnd.ms-excel')
part.set_payload(fp.read())
fp.close()
encoders.encode_base64(part)
part.add_header('Content-Disposition', 'attachment', filename='Name File Here')
msg.attach(part)
smtp = smtplib.SMTP('smtp.gmail.com')
smtp.ehlo()
smtp.starttls()
smtp.login(username,password)
smtp.sendmail(send_from, send_to.split(',') + msg['Cc'].split(','), msg.as_string())
smtp.quit()

Question 60

With my code you can send email attachments using gmail you will need to:

set your gmail address at “YOUR SMTP EMAIL HERE“

set your gmail account password at “YOUR SMTP PASSWORD HERE_“

In the ___EMAIL TO RECEIVE THE MESSAGE_ part you need to set the destination email address.

Alarm notification is the subject,

Someone has entered the room, picture attached is the body

[“/home/pi/webcam.jpg”] is an image attachment.

#!/usr/bin/env python
import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEBase import MIMEBase
from email.MIMEText import MIMEText
from email.Utils import COMMASPACE, formatdate
from email import Encoders
import os

USERNAME = "___YOUR SMTP EMAIL HERE___"
PASSWORD = "__YOUR SMTP PASSWORD HERE___"

def sendMail(to, subject, text, files=[]):
    assert type(to)==list
    assert type(files)==list

    msg = MIMEMultipart()
    msg['From'] = USERNAME
    msg['To'] = COMMASPACE.join(to)
    msg['Date'] = formatdate(localtime=True)
    msg['Subject'] = subject

    msg.attach( MIMEText(text) )

    for file in files:
        part = MIMEBase('application', "octet-stream")
        part.set_payload( open(file,"rb").read() )
        Encoders.encode_base64(part)
        part.add_header('Content-Disposition', 'attachment; filename="%s"'
                       % os.path.basename(file))
        msg.attach(part)

    server = smtplib.SMTP('smtp.gmail.com:587')
    server.ehlo_or_helo_if_needed()
    server.starttls()
    server.ehlo_or_helo_if_needed()
    server.login(USERNAME,PASSWORD)
    server.sendmail(USERNAME, to, msg.as_string())
    server.quit()

sendMail( ["___EMAIL TO RECEIVE THE MESSAGE__"],
        "Alarm notification",
        "Someone has entered the room, picture attached",
        ["/home/pi/webcam.jpg"] )

Question 61

You can also specify the type of attachment you want in your e-mail, as an example I used pdf:

def send_email_pdf_figs(path_to_pdf, subject, message, destination, password_path=None):
    ## credits: http://linuxcursor.com/python-programming/06-how-to-send-pdf-ppt-attachment-with-html-body-in-python-script
    from socket import gethostname
    #import email
    from email.mime.application import MIMEApplication
    from email.mime.multipart import MIMEMultipart
    from email.mime.text import MIMEText
    import smtplib
    import json

    server = smtplib.SMTP('smtp.gmail.com', 587)
    server.starttls()
    with open(password_path) as f:
        config = json.load(f)
        server.login('me@gmail.com', config['password'])
        # Craft message (obj)
        msg = MIMEMultipart()

        message = f'{message}\nSend from Hostname: {gethostname()}'
        msg['Subject'] = subject
        msg['From'] = 'me@gmail.com'
        msg['To'] = destination
        # Insert the text to the msg going by e-mail
        msg.attach(MIMEText(message, "plain"))
        # Attach the pdf to the msg going by e-mail
        with open(path_to_pdf, "rb") as f:
            #attach = email.mime.application.MIMEApplication(f.read(),_subtype="pdf")
            attach = MIMEApplication(f.read(),_subtype="pdf")
        attach.add_header('Content-Disposition','attachment',filename=str(path_to_pdf))
        msg.attach(attach)
        # send msg
        server.send_message(msg)

inspirations/credits to: http://linuxcursor.com/python-programming/06-how-to-send-pdf-ppt-attachment-with-html-body-in-python-script

Question 62

How can I send the HTML content in an email using Python? I can send simple text.

Question 63

From Python v2.7.14 documentation – 18.1.11. email: Examples:

Here’s an example of how to create an HTML message with an alternative plain text version:

#! /usr/bin/python

import smtplib

from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

# me == my email address
# you == recipient's email address
me = "my@email.com"
you = "your@email.com"

# Create message container - the correct MIME type is multipart/alternative.
msg = MIMEMultipart('alternative')
msg['Subject'] = "Link"
msg['From'] = me
msg['To'] = you

# Create the body of the message (a plain-text and an HTML version).
text = "Hi!\nHow are you?\nHere is the link you wanted:\nhttp://www.python.org"
html = """\
<html>
  <head></head>
  <body>
    <p>Hi!<br>
       How are you?<br>
       Here is the <a href="http://www.python.org">link</a> you wanted.
    </p>
  </body>
</html>
"""

# Record the MIME types of both parts - text/plain and text/html.
part1 = MIMEText(text, 'plain')
part2 = MIMEText(html, 'html')

# Attach parts into message container.
# According to RFC 2046, the last part of a multipart message, in this case
# the HTML message, is best and preferred.
msg.attach(part1)
msg.attach(part2)

# Send the message via local SMTP server.
s = smtplib.SMTP('localhost')
# sendmail function takes 3 arguments: sender's address, recipient's address
# and message to send - here it is sent as one string.
s.sendmail(me, you, msg.as_string())
s.quit()

Question 64

You might try using my mailer module.

from mailer import Mailer
from mailer import Message

message = Message(From="me@example.com",
                  To="you@example.com")
message.Subject = "An HTML Email"
message.Html = """<p>Hi!<br>
   How are you?<br>
   Here is the <a href="http://www.python.org">link</a> you wanted.</p>"""

sender = Mailer('smtp.example.com')
sender.send(message)

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你对OOP失望吗？