So I set up a GCP VM with Ubuntu, from where I want to send regular reports through my mail provider with a python script. The smtp port is 587, and to my understanding the port was formerly closed in GCP environments but should now be available.
My script looks like this:
import smtplib,ssl
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
from email.mime.text import MIMEText
from email.utils import formatdate
from email import encoders
import time
timestr = time.strftime('(%Y-%m-%d)')
username='me#mydomain.de'
password='secretpw'
send_from = 'me#mydomain.de'
send_to = 'recipient#herdomain.com'
body = 'hello'
msg = MIMEMultipart()
msg['From'] = send_from
msg['To'] = send_to
msg['Date'] = formatdate(localtime = True)
msg['Subject'] = 'important email'
msg.attach(MIMEText(body, 'plain'))
server = smtplib.SMTP('smtp.ionos.de')
port = '587'
smtp = smtplib.SMTP('smtp.ionos.de')
smtp.ehlo()
smtp.starttls()
smtp.login(username,password)
smtp.sendmail(send_from, send_to.split(','), msg.as_string())
smtp.quit()
On execution, the machine takes some time before outputting a timeout:
Traceback (most recent call last):
File "test.py", line 25, in <module>
server = smtplib.SMTP('smtp.ionos.de')
File "/usr/lib/python3.8/smtplib.py", line 255, in __init__
(code, msg) = self.connect(host, port)
File "/usr/lib/python3.8/smtplib.py", line 339, in connect
self.sock = self._get_socket(host, port, self.timeout)
File "/usr/lib/python3.8/smtplib.py", line 310, in _get_socket
return socket.create_connection((host, port), timeout,
File "/usr/lib/python3.8/socket.py", line 808, in create_connection
raise err
File "/usr/lib/python3.8/socket.py", line 796, in create_connection
sock.connect(sa)
TimeoutError: [Errno 110] Connection timed out
I can, however, ping smtp.ionos.de as well as telnet smtp.ionos.de 587 form the cl with result of a working ping and connection.
I also tried this with other email providers including gmail and get stuck with the exact same outcome.
Anyone? Help appreciated, thanks.
Your code has multiple problems:
Connecting to the server twice.
Not specifying the port number when connecting
Not creating an SSL context for encryption.
In your code, replace these lines:
server = smtplib.SMTP('smtp.ionos.de')
port = '587'
smtp = smtplib.SMTP('smtp.ionos.de')
smtp.ehlo()
smtp.starttls()
With:
context = ssl.create_default_context()
smtp = smtplib.SMTP('smtp.ionos.de', 587)
smtp.ehlo()
smtp.starttls(context=context)
Here is a full example tested with a GCP Cloud Function. No libs are needed in requirements.txt. This uses Python 3.9. The code uses an App Password because normal users require 2FA.
import smtplib
import ssl
def send_email(request):
"""Responds to any HTTP request.
Args:
request (flask.Request): HTTP request object.
Returns:
The response text or any set of values that can be turned into a
Response object using
`make_response <http://flask.pocoo.org/docs/1.0/api/#flask.Flask.make_response>`.
"""
gmail_user = 'user#gmail.com'
gmail_password = 'YOURPASSWORD'
sent_from = gmail_user
to = ['foo#bar.com']
subject = 'Test e-mail from Python'
body = 'Test e-mail body'
email_text = """\
From: %s
To: %s
Subject: %s
%s
""" % (sent_from, ", ".join(to), subject, body)
context = ssl.create_default_context()
server = smtplib.SMTP('smtp.gmail.com', 587)
server.ehlo()
server.starttls(context=context)
server.login(gmail_user, gmail_password)
server.sendmail(sent_from, to, body)
server.close()
print('Email sent (print)!')
return f'Email sent (return)!'
I am successfully able to send email using the smtplib module. But when the emial is sent, it does not include the subject in the email sent.
import smtplib
SERVER = <localhost>
FROM = <from-address>
TO = [<to-addres>]
SUBJECT = "Hello!"
message = "Test"
TEXT = "This message was sent with Python's smtplib."
server = smtplib.SMTP(SERVER)
server.sendmail(FROM, TO, message)
server.quit()
How should I write "server.sendmail" to include the SUBJECT as well in the email sent.
If I use, server.sendmail(FROM, TO, message, SUBJECT), it gives error about "smtplib.SMTPSenderRefused"
Attach it as a header:
message = 'Subject: {}\n\n{}'.format(SUBJECT, TEXT)
and then:
server = smtplib.SMTP(SERVER)
server.sendmail(FROM, TO, message)
server.quit()
Also consider using standard Python module email - it will help you a lot while composing emails. Using it would look like this:
from email.message import EmailMessage
msg = EmailMessage()
msg['Subject'] = SUBJECT
msg['From'] = FROM
msg['To'] = TO
msg.set_content(TEXT)
server.send_message(msg)
This will work with Gmail and Python 3.6+ using the new "EmailMessage" object:
import smtplib
from email.message import EmailMessage
msg = EmailMessage()
msg.set_content('This is my message')
msg['Subject'] = 'Subject'
msg['From'] = "me#gmail.com"
msg['To'] = "you#gmail.com"
# Send the message via our own SMTP server.
server = smtplib.SMTP_SSL('smtp.gmail.com', 465)
server.login("me#gmail.com", "password")
server.send_message(msg)
server.quit()
try this:
import smtplib
from email.mime.multipart import MIMEMultipart
msg = MIMEMultipart()
msg['From'] = 'sender_address'
msg['To'] = 'reciver_address'
msg['Subject'] = 'your_subject'
server = smtplib.SMTP('localhost')
server.sendmail('from_addr','to_addr',msg.as_string())
You should probably modify your code to something like this:
from smtplib import SMTP as smtp
from email.mime.text import MIMEText as text
s = smtp(server)
s.login(<mail-user>, <mail-pass>)
m = text(message)
m['Subject'] = 'Hello!'
m['From'] = <from-address>
m['To'] = <to-address>
s.sendmail(<from-address>, <to-address>, m.as_string())
Obviously, the <> variables need to be actual string values, or valid variables, I just filled them in as place holders. This works for me when sending messages with subjects.
See the note at the bottom of smtplib's documentation:
In general, you will want to use the email package’s features to construct an email message, which you can then convert to a string and send via sendmail(); see email: Examples.
Here's the link to the examples section of email's documentation, which indeed shows the creation of a message with a subject line. https://docs.python.org/3/library/email.examples.html
It appears that smtplib doesn't support subject addition directly and expects the msg to already be formatted with a subject, etc. That's where the email module comes in.
import smtplib
# creates SMTP session
List item
s = smtplib.SMTP('smtp.gmail.com', 587)
# start TLS for security
s.starttls()
# Authentication
s.login("login mail ID", "password")
# message to be sent
SUBJECT = "Subject"
TEXT = "Message body"
message = 'Subject: {}\n\n{}'.format(SUBJECT, TEXT)
# sending the mail
s.sendmail("from", "to", message)
# terminating the session
s.quit()
I think you have to include it in the message:
import smtplib
message = """From: From Person <from#fromdomain.com>
To: To Person <to#todomain.com>
MIME-Version: 1.0
Content-type: text/html
Subject: SMTP HTML e-mail test
This is an e-mail message to be sent in HTML format
<b>This is HTML message.</b>
<h1>This is headline.</h1>
"""
try:
smtpObj = smtplib.SMTP('localhost')
smtpObj.sendmail(sender, receivers, message)
print "Successfully sent email"
except SMTPException:
print "Error: unable to send email"
code from: http://www.tutorialspoint.com/python/python_sending_email.htm
In case of wrapping it in a function, this should work as a template.
def send_email(login, password, destinations, subject, message):
server = smtplib.SMTP_SSL("smtp.gmail.com", 465)
server.login(login, password)
message = 'Subject: {}\n\n{}'.format(subject, message)
for destination in destinations:
print("Sending email to:", destination)
server.sendmail(login, destinations, message)
server.quit()
try this out :
from = "myemail#site.com"
to= "someemail#site.com"
subject = "Hello there!"
body = "Have a good day."
message = "Subject:" + subject + "\n" + body
server.sendmail(from, , message)
server.quit()
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).