Setting a sender name when sending an email with smtplib and python - python

I am creating an endpoint where a user can get sent an email with a code to reset their password with.
When sending the email, I am using the built in smtplib that comes bundled with python.
When a request is made to the endpoint, I want the sender name to appear instead of the raw email. How can I accomplish this?
Here's my code:
code = ''.join(random.choice(string.ascii_letters + string.digits) for _ in range(8))
content = 'Subject: {}\n\n{}'.format('Your code to reset your password:', 'Your Code: {}'.format(code))
mail = smtplib.SMTP('smtp.gmail.com', 587)
mail.ehlo()
mail.starttls()
mail.login(os.environ.get('email'), os.environ.get('password'))
mail.sendmail(os.environ.get('email'), user.email, content)
mail.close()
When searching the web for similar questions, the only solution that came across was to add this line of code: content['From'] = "Your name "
Adding this line of code didn't send the email at all.
Does anybody know how to add the sender name to the email? Thank You.

Related

smtplib.SMTPConnectError: (421, b'Service not available') while using my pycharm to send a mail with python

i got the above error each time i try sending a mail using python. am somewhat new to this whole smtp thing. below is my code. please help.
import smtplib
my_email ="okoyekennethoptimizer#gmail.com"
password = "nmmrjdphptwfgjgb"
connection = smtplib.SMTP("smtp.gmail.com", 587)
connection.starttls()
connection.login(user=my_email, password=password)
connection.sendmail(from_addr=my_email, to_addrs="okoyek10#yahoomail.com", msg="hello")
connection.close()
please help. its been giving me tough time
Try creating an "App password" after enabling 2-step auth in your Gmail. You still need to do some lookup but generally, this should help.
The trick is in:
Enabling less secure apps.
Make sure your account has 2FA.
Generating an app password under My Google Account>>Security under Signing in to Google.
Replace the 16-digit password with the password you've placed in the password variable.
Just before connection.starttls() add connection.ehlo().
A code for example:
my_email ="okoyekennethoptimizer#gmail.com"
app_password = "16-digit-app-password"
connection = smtplib.SMTP("smtp.gmail.com", 587)
connection.ehlo()
connection.starttls()
connection.login(user=my_email, password=app_password)
Also you SHOULD create From:, To:, and Subject: message headers, separated from the message body by a blank line.
E.g.
msg = "\r\n".join([
"From: user_me#gmail.com",
"To: user_you#gmail.com",
"Subject: Any subject",
"",
"Good Evening, how are you?"
])

How to send an email using python after Google's policy update on not allowing just username and password?

I am trying to learn how to send an email using python. All the tutorials on the web that I have read explain how to do it using Gmail.
But, from 30/05/2022 (despite the fact that everybody is free to do whatever he wants with his account) Google has a new policy that states:
To help keep your account secure, starting May 30, 2022, Google will no longer support the use of third-party apps or devices that only ask for your username and password for you. Sign in to your Google account.
Source: https://support.google.com/accounts/answer/6010255
And we get:
So my question is there any other way to send an email using python, (including email accounts belonging to an other company)?
Here is my function to send an email:
def send_email_fct(filename, filepath, fromaddr, mdpfrom, toaddr):
"""" filename: file name to be sent with extension
filepath: file path of the file to be sent
fromaddr: sender email address
mdpfrom: password of sender email address
toaddr: receiver email address"""
msg = MIMEMultipart() # instance of MIMEMultipart
msg['From'] = fromaddr
msg['To'] = toaddr
msg['Subject'] = "data file"
body_email = "Body_of_the_mail"
msg.attach(MIMEText(body_email, 'plain'))
attachment = open(filepath, 'rb') # open the file to be sent
p = MIMEBase('application', 'octet-stream') # instance of MIMEBase
p.set_payload(attachment.read())
encoders.encode_base64(p)
p.add_header('Content-Disposition', "attachment; filename= %s" % filename)
msg.attach(p) # attach the instance 'p' to instance 'msg'
s = smtplib.SMTP('smtp.gmail.com', 587) # SMTP
s.starttls()
s.login(fromaddr, mdpfrom)
text = msg.as_string()
s.sendmail(from_email_addr, toaddr, text) # sending the email
s.quit() # terminating the session
And I get this error:
smtplib.SMTPAuthenticationError: (535, b'5.7.8 Username and Password not accepted. Learn more at\n5.7.8 https://support.google.com/mail/?p=BadCredentials c12-20020aa7d60c000000b0042be14040c1sm2612116edr.86 - gsmtp')
To fix this problem, I think that the only line that need to be change is this one:
s = smtplib.SMTP('smtp.gmail.com', 587)
If you know by what I can change it or if you see any other error, it will help me a lot! :-)
Here is a more precise answer with all the main steps. I hope it will help other people.
Log in into your email account: https://myaccount.google.com
Then go to the security part
Be sure that you have turn on two steps verification and click on "App password"
After select email and the corresponding device
It will generate a password that will look like this; it is this password that you have to use in your python script.
This password will appear here; in this place, you will have the option to erase it (so it will be no longer be useful to connect to your email account).
Hope it will help other people!
Solved it by creating App password. You must got to Google account. Security tab, active 2 Step Verification. After this new option under "Signing in to Google" the "App passwords" option will be actived. Just create one app password and use as password to authenticate

why do I get a smtplib.SMTPRecipientsRefused: Error?

So basically I am trying to develop a python program that lets me send emails to my friend's email without signing in. However I get a Error that says smtplib.SMTPRecipientsRefused:
Here is my code..
import smtplib
myemail = 'myemail#gmail.com'
mypassword = 'password'
friendemail = input('Please enter the email of the person:')
with smtplib.SMTP("smtp.gmail.com", 587) as smtp:
subject = input("Subject: ")
body = input('Message:')
msg = f"subject:{subject}\n\n{body}"
smtp.starttls()
smtp.login(myemail,mypassword)
smtp.sendmail(myemail, mypassword, msg)
It would be great if anyone could help! Thank YOU!
I tried to run the code you provided, and there was a problem with the image below, right?
According to the prompt, I changed the parameters of sendmail.
smtp.sendmail(myemail, friendemail, msg)
The parameters here represent: sender's mailbox, recipient's mailbox, and email content.
After the change, I successfully ran it in vscode and sent an email to my friend's Gmail email.
Note: the email name and password should be filled in correctly and can not be omitted.

Trying to send an Email through python with Gmail or Outlook

Hello I am fairly new to python and stumbled upon this cool feauture that with only a few lines of code, python can send emails.
I am currently not sure if the code that I have below works or if its something on my end? Since I am totally new to this I have no way to test if I am on the right track or not.
When running the code, it compiles with no problem but I never receive the messages.
Also, I am sending emails to myself, so I should see them rather quickly.
Here is my Outlook code:
import win32com.client as win32
outlook = win32.Dispatch('outlook.application')
mail = outlook.CreateItem(0)
mail.To = 'myemail#hotmail.com'
mail.Subject = 'Hello this is you!
mail.Body = 'Hello!!!!!!'
mail.HTMLBody = '<h2>This is an H2 message</h2>' #this field is optional
# To attach a file to the email (optional):
attachment = "C:/Users/OneDrive/Documents/Desktop/Social_Network_Ads.csv"
mail.Attachments.Add(attachment)
mail.Send()
Here is my Gmail Code:
import smtplib
fromaddr = 'myemail#gmail.com'
toaddrs = 'myemail#gmail.com'
msg = 'There was a terrible error that occured and I wanted you to know!'
# Credentials (if needed)
username = '###username###'
password = '###password###'
# The actual mail send
server = smtplib.SMTP('smtp.gmail.com:587')
server.starttls()
server.login(username,password)
server.sendmail(fromaddr, toaddrs, msg)
server.quit()
Please let me know why I am not receiving the email or why its showing as not sent?
EDIT:
I am connected to my local hotmail account and I am logged into Gmail, so I am hopping its not a connection issue.
When I go to check my sent folder nothing seems to have been sent

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).

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