Using Python smtplib to send to distribution lists - python

I'm using SMTPlib to automatically send out an email:
emailto = ['distro#email.com','me#email.com']
emailfrom = "me#email.com"
msg = MIMEMultipart('related')
msg['Subject'] = currentdate + " Subject"
msg['From'] = emailfrom
msg['To'] = ", ".join(emailto)
msgAlternative = MIMEMultipart('alternative')
msg.attach(msgAlternative)
msgAlternative.attach(msgText)
smtpObj = smtplib.SMTP('mail.email.com')
smtpObj.ehlo()
smtpObj.starttls()
smtpObj.sendmail(emailfrom, emailto, msg.as_string())
smtpObj.quit()
When I use this code I get the email, with distro#email.com in the "To:" line as well, but no one in distro#email.com gets it. I've sent to distribution lists before with no problem, but this specific one will not work. It is a fairly large list (~100 recipients)

Errors may pass unnoticed, because you are not checking the result of .sendmail() when sending to multiple addresses.
This method will not raise an exception if it succeeds to send the email to at least one recipient.
The important part of the docs is:
This method will return normally if the mail is accepted for at least
one recipient. Otherwise it will raise an exception.
If this method does not raise an exception, it returns a dictionary,
with one entry for each recipient that was refused. Each entry
contains a tuple of the SMTP error code and the accompanying error
message sent by the server.
Something like this should help to find the problem:
errors = smtpObj.sendmail(emailfrom, emailto, msg.as_string())
for recipient, (code, errmsg) in errors.items():
# ... print or log the error ...

Related

Efficiency of sending email in python

I am building a python program which can send emails in a certain time, below is my current code
def send_email(user):
msg = email.message.EmailMessage()
msg["From"] = "my Email"
msg["To"] = user
msg["Subject"] = "Subject"
msg.add_alternative("content", subtype = "html")
server = smtplib.SMTP_SSL("smtp.gmail.com", 465)
server.login("account", "password")
server.send_message(msg)
server.close()
I wanted to move the login process outside the send_email function so it doesn't have to login every time it needs to send emails, so I did this
server = smtplib.SMTP_SSL("smtp.gmail.com", 465)
server.login("account", "password")
def send_email(user):
msg = email.message.EmailMessage()
msg["From"] = "my Email"
msg["To"] = user
msg["Subject"] = "Subject"
msg.add_alternative("content", subtype = "html")
server.send_message(msg)
server.close()
But I started to get this error: smtplib.SMTPServerDisconnected: please run connect() first.
How to solve this problem? Or is there any other way to improve the efficiency?
You are calling server.close() every time you are calling send_email so it makes sense you'll need to connect/login again before being able to send another mail.
Move the server.close call to outside of the send_email function.
However, are you sure that keeping the session open is the best approach here? First I'd try to use a profiler to make sure the bottleneck is indeed in the login process and not somewhere else (ie the send process itself).

Detect bounced emails in Python smtplib

I'm trying to catch all emails that bounced when sending them via smtplib in Python. I looked at this similar post which suggested adding an exception catcher, but I noticed that my sendmail function doesn't throw any exceptions even for fake email addresses.
Here is my send_email function which uses smtplib.
def send_email(body, subject, recipients, sent_from="myEmail#server.com"):
msg = MIMEText(body)
msg['Subject'] = subject
msg['From'] = sent_from
msg['To'] = ", ".join(recipients)
s = smtplib.SMTP('mySmtpServer:Port')
try:
s.sendmail(msg['From'], recipients, msg.as_string())
except SMTPResponseException as e:
error_code = e.smtp_code
error_message = e.smtp_error
print("error_code: {}, error_message: {}".format(error_code, error_message))
s.quit()
Sample call:
send_email("Body-Test", "Subject-Test", ["fakejfdklsa#jfdlsaf.com"], "myemail#server.com")
Since I set the sender as myself, I am able to receive the email bounce report in my sender's inbox:
<fakejfdklsa#jfdlsaf.com>: Host or domain name not found. Name service error
for name=jfdlsaf.com type=A: Host not found
Final-Recipient: rfc822; fakejfdklsa#jfdlsaf.com
Original-Recipient: rfc822;fakejfdklsa#jfdlsaf.com
Action: failed
Status: 5.4.4
Diagnostic-Code: X-Postfix; Host or domain name not found. Name service error
for name=jfdlsaf.com type=A: Host not found
Is there a way to get the bounce message through Python?
import poplib
from email import parser
#breaks with if this is left out for some reason (MAXLINE is set too low by default.)
poplib._MAXLINE=20480
pop_conn = poplib.POP3_SSL('your pop server',port)
pop_conn.user(username)
pop_conn.pass_(password)
#Get messages from server:
messages = [pop_conn.retr(i) for i in range(1, len(pop_conn.list()[1]) + 1)]
# Concat message pieces:
messages = ["\n".join(mssg[1]) for mssg in messages]
#Parse message intom an email object:
messages = [parser.Parser().parsestr(mssg) for mssg in messages]
for message in messages:
if "Undeliverable" in message['subject']:
print message['subject']
for part in message.walk():
if part.get_content_type():
body = str(part.get_payload(decode=True))
bounced = re.findall('[a-z0-9-_\.]+#[a-z0-9-\.]+\.[a-z\.]{2,5}',body)
if bounced:
bounced = str(bounced[0].replace(username,''))
if bounced == '':
break
print bounced
Hope this helps. This will check the contents of the mailbox for any undeliverable reports and read the message to find the email address that bounced.It then prints the result
You can use this function for gmail, basically checking the inbox that whether there is any email from Mail Delivery Subsystem with particular Text in body
def is_email_bounced(email, app_password, email_in_question):
obj = imaplib.IMAP4_SSL('imap.gmail.com',993)
obj.login(email, app_password)
obj.select() # Default Inbox
typ, data = obj.search(None, f'(TEXT "{email_in_question}") (Subject "Delivery Status Notification (Failure)")')
try: data.remove(b'')
except: pass
if len(data) >= 1: return True
else: return False
run this function after sending the email, and check whether email got bounced or not
e.g.
send_email("Body-Test", "Subject-Test", ["fakejfdklsa#jfdlsaf.com"], "myemail#server.com")
if is_email_bounced("youremail#gmail.com", "app_password", "email_in_ques_404#gmail.com"):
print("Email Bounced")
Change email Subject "Delivery Status Notification (Failure)" as per your email service provider, for this code, I've considered email provider as gmail

Python SMTPlib: Message sending in from header

In Python I'm attempting to send a message via SMTPlib. However, the message is always sending the entire message in the from header, and I have no idea how to fix it. It wasn't doing it before, but now it's always doing it. Here is my code:
import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
def verify(email, verify_url):
msg = MIMEMultipart()
msg['From'] = 'pyhubverify#gmail.com\n'
msg['To'] = email + '\n'
msg['Subject'] = 'PyHub verification' + '\n'
body = """ Someone sent a PyHub verification email to this address! Here is the link:
www.xxxx.co/verify/{1}
Not you? Ignore this email.
""".format(email, verify_url)
msg.attach(MIMEText(body, 'plain'))
server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login('pyhubverify#gmail.com', 'xxxxxx')
print msg.as_string()
server.sendmail(msg['From'], [email], body)
server.close()
Is there anything wrong with it, and is there a way I can fix it?
This line is the issue:
server.sendmail(msg['From'], [email], body)
You could fix it with:
server.sendmail(msg['From'], [email], msg.as_string())
You were sending the body instead of the whole message; the SMTP protocol expects the message to start with the headers... hence you are seeing the body where the headers is supposed to be.
You also need to remove the newline characters. Per rfc2822 the line-feed characters are undesirable alone:
A message consists of header fields (collectively called "the header
of the message") followed, optionally, by a body. The header is a
sequence of lines of characters with special syntax as defined in
this standard. The body is simply a sequence of characters that
follows the header and is separated from the header by an empty line
(i.e., a line with nothing preceding the CRLF).
Please try the following:
msg = MIMEMultipart()
email = 'recipient#example.com'
verify_url = 'http://verify.example.com'
msg['From'] = 'pyhubverify#gmail.com'
msg['To'] = email
msg['Subject'] = 'PyHub verification'
body = """ Someone sent a PyHub verification email to this address! Here is the link:
www.xxxx.co/verify/{1}
Not you? Ignore this email.
""".format(email, verify_url)
msg.attach(MIMEText(body, 'plain'))
print msg.as_string()

Cannot send an email with python smtp

I am developing an application using python where I need to send a file through mail. I wrote a program to send the mail but dont know there's something wrong. The code is posted below. Please any one help me with this smtp library. Is there's anything i m missing? And also can someone please tell me what will be the host in smtp! I am using smtp.gmail.com.
Also can any one tell me how can i email a file (.csv file). Thanks for the help!
#!/usr/bin/python
import smtplib
sender = 'someone#yahoo.com'
receivers = ['someone#yahoo.com']
message = """From: From Person <someone#yahoo.com>
To: To Person <someone#yahoo.com>
Subject: SMTP e-mail test
This is a test e-mail message.
"""
try:
smtpObj = smtplib.SMTP('smtp.gmail.com')
smtpObj.sendmail(sender, receivers, message)
print "Successfully sent email"
except:
print "Error: unable to send email"
You aren't logging in. There are also a couple reasons you might not make it through including blocking by your ISP, gmail bouncing you if it can't get a reverse DNS on you, etc.
try:
smtpObj = smtplib.SMTP('smtp.gmail.com', 587) # or 465
smtpObj.ehlo()
smtpObj.starttls()
smtpObj.login(account, password)
smtpObj.sendmail(sender, receivers, message)
print "Successfully sent email"
except:
print "Error: unable to send email"
I just noticed your request to be able to attach a file. That changes things since now you need to deal with encoding. Still not that tough to follow though I don't think.
import os
import email
import email.encoders
import email.mime.text
import smtplib
# message/email details
my_email = 'myemail#gmail.com'
my_passw = 'asecret!'
recipients = ['jack#gmail.com', 'jill#gmail.com']
subject = 'This is an email'
message = 'This is the body of the email.'
file_name = 'C:\\temp\\test.txt'
# build the message
msg = email.MIMEMultipart.MIMEMultipart()
msg['From'] = my_email
msg['To'] = ', '.join(recipients)
msg['Date'] = email.Utils.formatdate(localtime=True)
msg['Subject'] = subject
msg.attach(email.MIMEText.MIMEText(message))
# build the attachment
att = email.MIMEBase.MIMEBase('application', 'octet-stream')
att.set_payload(open(file_name, 'rb').read())
email.Encoders.encode_base64(att)
att.add_header('Content-Disposition', 'attachment; filename="%s"' % os.path.basename(file_name))
msg.attach(att)
# send the message
srv = smtplib.SMTP('smtp.gmail.com', 587)
srv.ehlo()
srv.starttls()
srv.login(my_email, my_passw)
srv.sendmail(my_email, recipients, msg.as_string())

How to send an email with Python?

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?
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.
When I need to mail in Python, I use the mailgun API which gets a lot of the headaches with sending mails sorted out. They have a wonderful 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'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.
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
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()
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).
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)
It's probably putting tabs into your message. Print out message before you pass it to sendMail.
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)
I haven't been satisfied with the package options for sending emails and I decided to make and open source my own email sender. It is easy to use and capable of advanced use cases.
To install:
pip install redmail
Usage:
from redmail import EmailSender
email = EmailSender(
host="<SMTP HOST ADDRESS>",
port=<PORT NUMBER>,
)
email.send(
sender="me#example.com",
receivers=["you#example.com"],
subject="An example email",
text="Hi, this is text body.",
html="<h1>Hi,</h1><p>this is HTML body</p>"
)
If your server requires a user and a password, just pass user_name and password to the EmailSender.
I have included a lot of features wrapped in the send method:
Include attachments
Include images directly to the HTML body
Jinja templating
Prettier HTML tables out of the box
Documentation:
https://red-mail.readthedocs.io/en/latest/
Source code: https://github.com/Miksus/red-mail
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()
Another implementation using gmail let's say:
import smtplib
def send_email(email_address: str, subject: str, body: str):
"""
send_email sends an email to the email address specified in the
argument.
Parameters
----------
email_address: email address of the recipient
subject: subject of the email
body: body of the email
"""
server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login("email_address", "password")
server.sendmail("email_address", email_address,
"Subject: {}\n\n{}".format(subject, body))
server.quit()
I wrote a simple function send_email() for email sending with smtplib and email packages (link to my article). It additionally uses dotenv package to loads the sender email and password (please don't keep secrets in the code!). I was using Gmail for email service. The password was the App Password (here is Google docs on how to generate App Password).
import os
import smtplib
from email.message import EmailMessage
from dotenv import load_dotenv
_ = load_dotenv()
def send_email(to, subject, message):
try:
email_address = os.environ.get("EMAIL_ADDRESS")
email_password = os.environ.get("EMAIL_PASSWORD")
if email_address is None or email_password is None:
# no email address or password
# something is not configured properly
print("Did you set email address and password correctly?")
return False
# create email
msg = EmailMessage()
msg['Subject'] = subject
msg['From'] = email_address
msg['To'] = to
msg.set_content(message)
# send email
with smtplib.SMTP_SSL('smtp.gmail.com', 465) as smtp:
smtp.login(email_address, email_password)
smtp.send_message(msg)
return True
except Exception as e:
print("Problem during send email")
print(str(e))
return False
The above approach is OK for simple email sending. If you are looking for more advanced features, such as HTML content or attachments - it, of course, can be hand-coded, but I would recommend using existing packages, for example yagmail.
Gmail has a limit of 500 emails per day. For sending many emails per day please consider transactional email service providers, like Amazon SES, MailGun, MailJet, or SendGrid.
import smtplib
s = smtplib.SMTP(your smtp server, smtp port) #SMTP session
message = "Hii!!!"
s.sendmail("sender", "Receiver", message) # sending the mail
s.quit() # terminating the session
just to complement the answer and so that your mail delivery system can be scalable.
I recommend having a configuration file (it can be .json, .yml, .ini, etc) with the sender's email configuration , password and recipients.
This way you can create different customizable items according to your needs.
Below is a small example with 3 files, config, functions and main. Text-only mailing.
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"]
These items will be called from main.py, which will return their respective values.
File with functions functions_email.py:
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()
File 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)
Best regards!
import smtplib, ssl
port = 587 # For starttls
smtp_server = "smtp.office365.com"
sender_email = "170111018#student.mit.edu.tr"
receiver_email = "professordave#hotmail.com"
password = "12345678"
message = """\
Subject: Final exam
Teacher when is the final exam?"""
def SendMailf():
context = ssl.create_default_context()
with smtplib.SMTP(smtp_server, port) as server:
server.ehlo() # Can be omitted
server.starttls(context=context)
server.ehlo() # Can be omitted
server.login(sender_email, password)
server.sendmail(sender_email, receiver_email, message)
print("mail send")
After a lot of fiddling with the examples e.g here
this now works for me:
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")
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).

Categories