Related
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).
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).
I want to write a function to attach a csv file to an E-Mail with HTML text. Everything works fine. I sends the E-Mail with the text and the attachment with the right information. Only the data format is wrong. I tried different varieties in MIMEBASE('application', 'octet-stream'). This one gives me a '.bin' data which I cannot open on macOS. Others gave me a plain text data which included the right data, but I don't wanna copy it manually into a csv. Does someone have a solution how I get out the data as '.csv'? Most of the solutions I found here just looked like my code below.
Code:
def send_mail(receiver, subject, filepath, attachname):
#login information
port = 123
smtp_server = "test.server.com"
login = "testlogin" # your login
password = "12345678910" # your password
# specify the sender’s and receiver’s email addresses and text in HTML
sender = "testmail#test.com"
receiver = receiver
message = MIMEMultipart()
message["Subject"] = subject
message["From"] = sender
message["To"] = COMMASPACE.join([receiver])
html_text = """\
<html>
<body>
<p>Hi ,<br>
<p>kind regards</p>
<p> This is an automatized Mail </p>
</body>
</html>
"""
# convert both parts to MIMEText objects and add them to the MIMEMultipart message
text = MIMEText(html_text, "html")
message.attach(text)
#add a file as attachment
file = open(filepath, "rb")
att = MIMEBase('application', 'octet-stream')
att.set_payload(file.read())
encoders.encode_base64(att)
att.add_header("Content_Disposition",
'attachment', filename = attachname)
message.attach(att)
try:
#send your message with credentials specified above
with smtplib.SMTP(smtp_server, port) as server:
server.connect(smtp_server, port)
server.ehlo()
server.starttls()
server.ehlo()
server.login(login, password)
server.sendmail(sender, receiver, message.as_string())
# tell the script to report if your message was sent or which errors need to be fixed
print('Sent')
except (gaierror, ConnectionRefusedError):
print('Failed to connect to the server. Bad connection settings?')
except smtplib.SMTPServerDisconnected:
print('Failed to connect to the server. Wrong user/password?')
except smtplib.SMTPException as e:
print('SMTP error occurred: ' + str(e))
send_mail('testmail#test.com', 'TEST', '/Users/Changer/Desktop/test.csv', 'test.csv')
Email clients may use the content type of the attachment part to determine which program is used to open the attachment, so specifying a suitable content type may help.
This code uses the standard library's mimetypes module to try to guess the correct content type of the attachment, based on its name:
import mimetypes
mimetypes.init()
def send_mail(receiver, subject, filepath, attachname):
...
mimetype, _ = mimetypes.guess_type(attachname)
if mimetype is None:
mimetype = 'application/octet-stream'
type_, _, subtype = mimetype.partition('/')
att = MIMEBase(type_, subtype)
...
The above code will generate these headers for the attachment:
b'Content-Type: text/x-comma-separated-values'
b'MIME-Version: 1.0'
b'Content-Transfer-Encoding: base64'
b'Content_Disposition: attachment; filename="test.csv"'
Whether a mac will open a csv with a spreadsheet application (I assume this is what you want) depends on the configuration of the machine (See this blog for an example). If you want to be sure open the file in a spreadsheet it might be more effective to convert it to a spreadsheet format and send the spreadsheet.
I have an output of df['sentiments'] as sentiment output of twitter tweets. I want to mail that sentiment output over email to another person automatically.
Store it in a file using df['sentiments'].to_csv('sentiment.csv'). And send the file sentiment.csv via mail.
More details about to_csv function can be found here in the official docs.
EDIT: Expanding the answer for the sake of making it less troublesome. Credits for this part
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders
mail_content = '''Sample message'''
#The mail addresses and password
sender_address = 'sender#gmail.com'
sender_pass = 'xxxxxxxxxxx'
receiver_address = 'receiver#gmail.com'
#Setup the MIME
message = MIMEMultipart()
message['From'] = sender_address
message['To'] = receiver_address
message['Subject'] = 'A test mail sent by Python. It has an attachment.'
#The body and the attachments for the mail
message.attach(MIMEText(mail_content, 'plain'))
filename = "database.txt" #- Attach the sentiment.csv and metadata file here
# Open PDF file in binary mode
with open(filename, "rb") as attachment:
# Add file as application/octet-stream
# Email client can usually download this automatically as attachment
part = MIMEBase("application", "octet-stream")
part.set_payload(attachment.read())
# Encode file in ASCII characters to send by email
encoders.encode_base64(part)
# Add header as key/value pair to attachment part
part.add_header(
"Content-Disposition",
f"attachment; filename= {filename}",
)
# Add attachment to message and convert message to string
message.attach(part)
#Create SMTP session for sending the mail
session = smtplib.SMTP('smtp.gmail.com', 587) #use gmail with port
session.ehlo()
session.starttls() #enable security
session.login(sender_address, sender_pass) #login with mail_id and password
text = message.as_string()
session.sendmail(sender_address, receiver_address, text)
session.quit()
print('Mail Sent')
If you get an error as follows:
(535, b'5.7.8 Username and Password not accepted. Learn more at\n5.7.8 https://support.google.com/mail/?p=BadCredentials y186sm1525057pfy.66 - gsmtp')
Follow this thread
You can use the python built in library smtplib:
https://docs.python.org/3/library/smtplib.html to send emails.
You can use, for example the GMAIL mail server to send it.
https://www.tutorialspoint.com/send-mail-from-your-gmail-account-using-python
I was trying to get a python program send an attachment via Gmail.
I used the sample code found:
Sending email via gmail & python
The problem is when I sent Excel files like .xls, .xlsx, .xlsm to myself, I cannot open the attachments as they were corrupted, even though the original files are fine. But sending .csv works fine. The entire process does not pop up any warnings or error.
Question is: did oauth2.0 or Gmail or MIME mess up the attachment for some formats? And how can I tell the program upload and send attachment without modifying it?
Had similar issue.
The error is probably in encoding file into bytes.
My gmail still sends corrupted file, when I do want to send xlsx, but I managed to get correct email with xls format.
This is my code:
def create_message_with_excel_attachment(sender, to, subject, message_text, file):
message = MIMEMultipart()
message['to'] = to
message['from'] = sender
message['subject'] = subject
msg = MIMEText(message_text)
message.attach(msg)
part = MIMEBase('application', "vnd.ms-excel")
part.set_payload(open(file, "rb").read())
encoders.encode_base64(part)
part.add_header('Content-Disposition', 'attachment', filename=file)
message.attach(part)
raw = base64.urlsafe_b64encode(message.as_bytes())
raw = raw.decode()
return {'raw': raw}
def send_message(service, user_id, message):
message = (service.users().messages().send(userId=user_id, body=message).execute())
try:
print ('Message Id: %s' % message['id'])
return message
except:
print ('An error occurred:')