CODEWITHSUNDEEP
pythonsmtpgmail
Shaggy89
Shaggy89

Reputation: 709

Connection Error SMTP python: smtplib.SMTPServerDisconnected: please run connect() first

I've been using python for a bit now and have been using the email function without any errors in the past but on the latest program I have made I've been getting this error

 Traceback (most recent call last):
    File "daemon.py", line 62, in <module>
    scraper.run()
    File "c:\cfsresd\scraper.py", line 48, in run
    self.scrape()
    File "c:\cfsresd\scraper.py", line 44, in scrape
    handler(msg)
    File "daemon.py", line 57, in handler
    server.ehlo()
    File "C:\Python27\lib\smtplib.py", line 385, in ehlo
    self.putcmd(self.ehlo_msg, name or self.local_hostname)
    File "C:\Python27\lib\smtplib.py", line 318, in putcmd
    self.send(str) 
    File "C:\Python27\lib\smtplib.py", line 310, in send
    raise SMTPServerDisconnected('please run connect() first')
    smtplib.SMTPServerDisconnected: please run connect() first

I used the same email code for all my projects but this is first time is done it. I've tried adding the connect() but that made no difference. Below is email section of my script

msg = MIMEText ('%s - %s' % (msg.text, msg.channel))
    server = smtplib.SMTP('smtp.gmail.com:587')
    server.ehlo()
    server.starttls()
    msg['Subject'] = "msg.channel"
    msg['From'] = ('removed')
    msg['To'] = ('removed')
    server.login('user','password')
    server.sendmail(msg.get('From'),msg["To"],msg.as_string())
    server.close()
    server.ehlo()
    server.quit()
    print('sent')

Upvotes: 14

Views: 52109

Answers (10)

Shaggy89
Shaggy89

Reputation: 709

All sorted took a few idea and tried the code below:

msg = MIMEText ('%s - %s' % (msg.text, msg.channel))
server = smtplib.SMTP('smtp.gmail.com')
server.starttls()
server.login('user','pass')
msg['Subject'] = "msg.channel"
msg['From'] = ('from')
msg['To'] = ('to')
server.sendmail(msg.get('From'),msg["To"],msg.as_string())
server.quit()

So I removed ehlo(), close() and port number. Now I have to workout how to change the subject to msg.channel so it changes each time.

Upvotes: 8

Prashant Singh
Prashant Singh

Reputation: 1

Believe me or not solution is very simple....

Simply just don't close the connection our in other words don't quit the server.

remove "server.quit()".

now you'll be able to send as many mails you need to.

Upvotes: 0

yusuf yahya
yusuf yahya

Reputation: 1

raise SMTPServerDisconnected('please run connect() first')

if you had this error you my be want install this :

pip install django-smtp-ssl

this one to install smtp library and ssl protocol

its work perfecly for me

Upvotes: 0

Omid Estaji
Omid Estaji

Reputation: 320

I solved this error just by removing this line:

server.quit()

Upvotes: 0

Michał Tołkacz
Michał Tołkacz

Reputation: 131

I had a similar problem when I tried to send an e-mail from Celery (as a Docker container). I added env_file to the worker and beat containers in a docker compose file.

env_file: ./env/dev/.env

In that file I have an e-mail configuration.

EMAIL_HOST=smtp.gmail.com
EMAIL_HOST_USER=your_mail
EMAIL_HOST_PASSWORD=your_password
EMAIL_PORT=587

Upvotes: 0

DeepSpace
DeepSpace

Reputation: 81684

Try using SMTP's empty constructor, then call connect(host, port):

    server = smtplib.SMTP()
    server.connect('smtp.gmail.com', '587')
    server.ehlo()
    server.starttls()
    server.login(username, password)

Upvotes: 3

0x Tps
0x Tps

Reputation: 165

For Pyhton 3.6.*
Note : In gmail it will work only if 2-Step verification is turned off. Allow gmail to open via low secured app.

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

from_addr = 'sender-email-id'
to_addr = 'receiver-email-id'
text = 'Hi Friend!!!'

username = 'sender-username'
password = 'password'

msg = MIMEMultipart()

msg['From'] = from_addr
msg['To'] = to_addr
msg['Subject'] = 'Test Mail'
msg.attach(MIMEText(text))


server = smtplib.SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()
server.ehlo()
server.login(username,password)
server.sendmail(from_addr,to_addr,msg.as_string())
server.quit()

Upvotes: 1

PascalVKooten
PascalVKooten

Reputation: 21471

I'm the maintainer of yagmail, a package that should make it really easy to send an email.

import yagmail
yag = yagmail.SMTP('user','password')
yag.send(to = '[email protected]', subject = 'msg.channel')

when yag leaves scope, it will auto-close.

I would also advise you to register in keyring once, so you'll never have to write the password in a script. Just run once:

yagmail.register('user', 'password')

You can then shorten it to this:

SMTP().send('[email protected]', 'msg.channel')

You can install it with pip or pip3 (for Python 3). You can also read more about it, with functionality as easily adding attachments, inline images/html, aliases etc.

Upvotes: 0

dcg
dcg

Reputation: 4219

You can still have an encrypted connection with the smtp server by using the SMTP_SSL class without needing the starttls call (shorter). You don't need to be calling the ehlo every time, that's done automatically when needed, and when connecting to the default port, don't have to supply one when creating instances SMTP* classes.

msg = MIMEText ('%s - %s' % (msg.text, msg.channel))
msg['To'] = ','.join(receivers)
msg['Subject'] = 'msg.channel'
msg['From'] = '[email protected]'

Using SMTP with the starttls:

server = smtplib.SMTP('smtp.gmail.com')
server.starttls()
server.login('user', 'password')
server.sendmail(msg['From'], receivers, msg.as_string())

and now with the SMTP_SSL class

server = smtplib.SMTP_SSL('smtp.gmail.com')
server.login('user', 'password')
server.sendmail(msg['From'], receivers, msg.as_string())

and finally 

server.quit()

Upvotes: 2

Eric Renouf
Eric Renouf

Reputation: 14520

You have an ehlo after close. That seems unlikely to ever succeed. Also, quit does close so you can probably just get rid of the ehlo and close calls near the end

Upvotes: 1

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