Connection Error SMTP python

Question

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'

cheers for any help

shaggy

Solution 1

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.

thanks all

Solution 2

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)

Solution 3

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

Solution 4

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()

Solution 5

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()

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