Problem occured while sending mail

[sui]

this is my code

import sys, os, glob, datetime, time
import smtplib
## Parameters for SMTP session
port=587
SMTPserver= 'smtp.gmail.com'
SMTPuser= '[e-mail]'
pw= 'fill in here'
SENDER= SMTPuser

## Message details
FROM= SENDER
TO= '[e-mail]'
CC=FROM
##RECEIVERS= (TO, CC) ##proper way to send to both TO and CC
RECEIVERS= (TO,) ## ignore the CC address

subject= 'Test 1a'
message='*** Email test *** '

print 'Starting SMTP mail session on %s as %s ' %
(SMTPserver,SMTPuser)
session = smtplib.SMTP(SMTPserver,port)
session.set_debuglevel(0) # set debug level to 1 to see details
session.ehlo(SMTPuser) # say hello
session.starttls() # TLS needed
session.ehlo(SMTPuser) # say hello again, not sure why
session.login(SMTPuser, pw)

##Create HEADER + MESSAGE
HEADER= 'From: %s\r\n' % FROM
HEADER= HEADER + 'To: %s\r\n' % TO
HEADER= HEADER + 'Cc: %s\r\n' % CC
HEADER= HEADER + 'Subject: %s\r\n' % subject
BODY= HEADER + '\r\n' + message
print BODY

SMTPresult = session.sendmail(SENDER, RECEIVERS, BODY) ## send email

session.close()

Now when i run this .py file...as python mail.py
i can see only statement
starting smtp mail......n details
then nothing on screen after few minutes or after pressing ctrl +c
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 301, in connect
self.sock.connect(sa)
File "<string>", line 1, in connect
or may be conncetion time out

wats the solution for this

[sui]

On Sep 17, 5:04 pm, sui <[e-mail]:

this is my code

import sys, os, glob, datetime, time
import smtplib
## Parameters for SMTP session
port=587
SMTPserver= 'smtp.gmail.com'
SMTPuser= '[e-mail]'
pw= 'fill in here'
SENDER= SMTPuser

## Message details
FROM= SENDER
TO= '[e-mail]'
CC=FROM
##RECEIVERS= (TO, CC) ##proper way to send to both TO and CC
RECEIVERS= (TO,) ## ignore the CC address

subject= 'Test 1a'
message='*** Email test *** '

print 'Starting SMTP mail session on %s as %s ' %
(SMTPserver,SMTPuser)
session = smtplib.SMTP(SMTPserver,port)
session.set_debuglevel(0) # set debug level to 1 to see details
session.ehlo(SMTPuser) # say hello
session.starttls() # TLS needed
session.ehlo(SMTPuser) # say hello again, not sure why
session.login(SMTPuser, pw)

##Create HEADER + MESSAGE
HEADER= 'From: %s\r\n' % FROM
HEADER= HEADER + 'To: %s\r\n' % TO
HEADER= HEADER + 'Cc: %s\r\n' % CC
HEADER= HEADER + 'Subject: %s\r\n' % subject
BODY= HEADER + '\r\n' + message
print BODY

SMTPresult = session.sendmail(SENDER, RECEIVERS, BODY) ## send email

session.close()

Now when i run this .py file...as python mail.py
i can see only statement
starting smtp mail......n details
then nothing on screen after few minutes or after pressing ctrl +c
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 301, in connect
self.sock.connect(sa)
File "<string>", line 1, in connect
or may be conncetion time out

wats the solution for this

if i dont press cntrl + c then it shows
Starting SMTP mail session on smtp.gmail.com as go*********@gmail.com
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 310, in connect
raise socket.error, msg
socket.error: (110, 'Connection timed out')

plz help me its urgent.....i want to complete it as early as possible

[Peter Pearson]

On Wed, 17 Sep 2008 05:28:05 -0700 (PDT), sui <go*********@gmail.comwrote:

On Sep 17, 5:04 pm, sui <[e-mail]:

>this is my code

import sys, os, glob, datetime, time
import smtplib
## Parameters for SMTP session
port=587
SMTPserver= 'smtp.gmail.com'
SMTPuser= '[e-mail]'
pw= 'fill in here'
SENDER= SMTPuser

## Message details
FROM= SENDER
TO= '[e-mail]'
CC=FROM
##RECEIVERS= (TO, CC) ##proper way to send to both TO and CC
RECEIVERS= (TO,) ## ignore the CC address

subject= 'Test 1a'
message='*** Email test *** '

print 'Starting SMTP mail session on %s as %s ' %
(SMTPserver,SMTPuser)
session = smtplib.SMTP(SMTPserver,port)
session.set_debuglevel(0) # set debug level to 1 to see details
session.ehlo(SMTPuser) # say hello
session.starttls() # TLS needed
session.ehlo(SMTPuser) # say hello again, not sure why
session.login(SMTPuser, pw)

##Create HEADER + MESSAGE
HEADER= 'From: %s\r\n' % FROM
HEADER= HEADER + 'To: %s\r\n' % TO
HEADER= HEADER + 'Cc: %s\r\n' % CC
HEADER= HEADER + 'Subject: %s\r\n' % subject
BODY= HEADER + '\r\n' + message
print BODY

SMTPresult = session.sendmail(SENDER, RECEIVERS, BODY) ## send email

session.close()

Now when i run this .py file...as python mail.py
i can see only statement
starting smtp mail......n details
then nothing on screen after few minutes or after pressing ctrl +c
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 301, in connect
self.sock.connect(sa)
File "<string>", line 1, in connect
or may be conncetion time out

wats the solution for this

if i dont press cntrl + c then it shows
Starting SMTP mail session on smtp.gmail.com as go*********@gmail.com
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 310, in connect
raise socket.error, msg
socket.error: (110, 'Connection timed out')

plz help me its urgent.....i want to complete it as early as possible

I pasted your code into a file named temp.py,
and (perhaps superstitiously) added a backslash to this line:

>print 'Starting SMTP mail session on %s as %s ' %

Here's what it does (long line wrapped manually):

peter@eleodes:~$ python temp.py
Starting SMTP mail session on smtp.gmail.com as [e-mail]
Traceback (most recent call last):
File "temp.py", line 27, in ?
session.login(SMTPuser, pw)
File "/usr/lib/python2.4/smtplib.py", line 591, in login
raise SMTPAuthenticationError(code, resp)
smtplib.SMTPAuthenticationError: (535, \
'5.7.1 Username and Password not accepted. Learn more at\n' \
'5.7.1 http://mail.google.com/support/bin/answer.py?answer=' \
'14257 a8sm34686663poa.12')
peter@eleodes:~$

This indicates that it got much farther than when you ran it, since
your timeout message comes from the smtplib.SMTP call several lines
before the session.login call.

As a simple connectivity test, you might see whether you can connect
using telnet:

peter@eleodes:~$ telnet smtp.gmail.com 587
Trying 72.14.253.109...
Connected to gmail-smtp.l.google.com.
Escape character is '^]'.
220 mx.google.com ESMTP m27sm34789033pof.6
^]c

telnetc
Connection closed.
peter@eleodes:~$

--
To email me, substitute nowhere->spamcop, invalid->net.

[sui]

On Sep 17, 8:04 pm, Peter Pearson <[e-mail]:

On Wed, 17 Sep 2008 05:28:05 -0700 (PDT), sui <[e-mail]:

On Sep 17, 5:04 pm, sui <[e-mail]:

this is my code

import sys, os, glob, datetime, time
import smtplib
## Parameters for SMTP session
port=587
SMTPserver= 'smtp.gmail.com'
SMTPuser= '[e-mail]'
pw= 'fill in here'
SENDER= SMTPuser

## Message details
FROM= SENDER
TO= '[e-mail]'
CC=FROM
##RECEIVERS= (TO, CC) ##proper way to send to both TO and CC
RECEIVERS= (TO,) ## ignore the CC address

subject= 'Test 1a'
message='*** Email test *** '

print 'Starting SMTP mail session on %s as %s ' %
(SMTPserver,SMTPuser)
session = smtplib.SMTP(SMTPserver,port)
session.set_debuglevel(0) # set debug level to 1 to see details
session.ehlo(SMTPuser) # say hello
session.starttls() # TLS needed
session.ehlo(SMTPuser) # say hello again, not sure why
session.login(SMTPuser, pw)

##Create HEADER + MESSAGE
HEADER= 'From: %s\r\n' % FROM
HEADER= HEADER + 'To: %s\r\n' % TO
HEADER= HEADER + 'Cc: %s\r\n' % CC
HEADER= HEADER + 'Subject: %s\r\n' % subject
BODY= HEADER + '\r\n' + message
print BODY

SMTPresult = session.sendmail(SENDER, RECEIVERS, BODY) ## send email

session.close()

Now when i run this .py file...as python mail.py
i can see only statement
starting smtp mail......n details
then nothing on screen after few minutes or after pressing ctrl +c
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 301, in connect
self.sock.connect(sa)
File "<string>", line 1, in connect
or may be conncetion time out

wats the solution for this

if i dont press cntrl + c then it shows
Starting SMTP mail session on smtp.gmail.com as [e-mail]
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 310, in connect
raise socket.error, msg
socket.error: (110, 'Connection timed out')

plz help me its urgent.....i want to complete it as early as possible

I pasted your code into a file named temp.py,
and (perhaps superstitiously) added a backslash to this line:

print 'Starting SMTP mail session on %s as %s ' %

Here's what it does (long line wrapped manually):

peter@eleodes:~$ python temp.py
Starting SMTP mail session on smtp.gmail.com as [e-mail]
Traceback (most recent call last):
File "temp.py", line 27, in ?
session.login(SMTPuser, pw)
File "/usr/lib/python2.4/smtplib.py", line 591, in login
raise SMTPAuthenticationError(code, resp)
smtplib.SMTPAuthenticationError: (535, \
'5.7.1 Username and Password not accepted. Learn more at\n' \
'5.7.1http://mail.google.com/support/bin/answer.py?answer='\
'14257 a8sm34686663poa.12')
peter@eleodes:~$

This indicates that it got much farther than when you ran it, since
your timeout message comes from the smtplib.SMTP call several lines
before the session.login call.

As a simple connectivity test, you might see whether you can connect
using telnet:

peter@eleodes:~$ telnet smtp.gmail.com 587
Trying 72.14.253.109...
Connected to gmail-smtp.l.google.com.
Escape character is '^]'.
220 mx.google.com ESMTP m27sm34789033pof.6
^]c

telnetc
Connection closed.
peter@eleodes:~$

--
To email me, substitute nowhere->spamcop, invalid->net.

Actually i m working at place where proxy server has been
working......so is it problem caused by that proxy server..peter thnks
for suggestion but still its not working......

[sui]

On Sep 17, 8:04 pm, Peter Pearson <[e-mail]:

On Wed, 17 Sep 2008 05:28:05 -0700 (PDT), sui <[e-mail]:

On Sep 17, 5:04 pm, sui <[e-mail]:

this is my code

import sys, os, glob, datetime, time
import smtplib
## Parameters for SMTP session
port=587
SMTPserver= 'smtp.gmail.com'
SMTPuser= '[e-mail]'
pw= 'fill in here'
SENDER= SMTPuser

## Message details
FROM= SENDER
TO= '[e-mail]'
CC=FROM
##RECEIVERS= (TO, CC) ##proper way to send to both TO and CC
RECEIVERS= (TO,) ## ignore the CC address

subject= 'Test 1a'
message='*** Email test *** '

print 'Starting SMTPmailsession on %s as %s ' %
(SMTPserver,SMTPuser)
session = smtplib.SMTP(SMTPserver,port)
session.set_debuglevel(0) # set debug level to 1 to see details
session.ehlo(SMTPuser) # say hello
session.starttls() # TLS needed
session.ehlo(SMTPuser) # say hello again, not sure why
session.login(SMTPuser, pw)

##Create HEADER + MESSAGE
HEADER= 'From: %s\r\n' % FROM
HEADER= HEADER + 'To: %s\r\n' % TO
HEADER= HEADER + 'Cc: %s\r\n' % CC
HEADER= HEADER + 'Subject: %s\r\n' % subject
BODY= HEADER + '\r\n' + message
print BODY

SMTPresult = session.sendmail(SENDER, RECEIVERS, BODY) ## send email

session.close()

Now when i run this .py file...as pythonmail.py
i can see only statement
starting smtpmail......n details
then nothing on screen after few minutes or after pressing ctrl +c
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 301, in connect
self.sock.connect(sa)
File "<string>", line 1, in connect
or may be conncetion time out

wats the solution for this

if i dont press cntrl + c then it shows
Starting SMTPmailsession on smtp.gmail.com as [e-mail]
Traceback (most recent call last):
File "mail4.py", line 21, in <module>
session = smtplib.SMTP(SMTPserver,port)
File "/usr/local/lib/python2.5/smtplib.py", line 244, in __init__
(code, msg) = self.connect(host, port)
File "/usr/local/lib/python2.5/smtplib.py", line 310, in connect
raise socket.error, msg
socket.error: (110, 'Connection timed out')

plz help me its urgent.....i want to complete it as early as possible

I pasted your code into a file named temp.py,
and (perhaps superstitiously) added a backslash to this line:

print 'Starting SMTPmailsession on %s as %s ' %

Here's what it does (long line wrapped manually):

peter@eleodes:~$ python temp.py
Starting SMTPmailsession on smtp.gmail.com as [e-mail]
Traceback (most recent call last):
File "temp.py", line 27, in ?
session.login(SMTPuser, pw)
File "/usr/lib/python2.4/smtplib.py", line 591, in login
raise SMTPAuthenticationError(code, resp)
smtplib.SMTPAuthenticationError: (535, \
'5.7.1 Username and Password not accepted. Learn more at\n' \
'5.7.1http://mail.google.com/support/bin/answer.py?answer='\
'14257 a8sm34686663poa.12')
peter@eleodes:~$

This indicates that it got much farther than when you ran it, since
your timeout message comes from the smtplib.SMTP call several lines
before the session.login call.

As a simple connectivity test, you might see whether you can connect
using telnet:

peter@eleodes:~$ telnet smtp.gmail.com 587
Trying 72.14.253.109...
Connected to gmail-smtp.l.google.com.
Escape character is '^]'.
220 mx.google.com ESMTP m27sm34789033pof.6
^]c

telnetc
Connection closed.
peter@eleodes:~$

--
To email me, substitute nowhere->spamcop, invalid->net.

even i couldnt connect using telnet....
msg comes host is down

[Peter Pearson]

On Wed, 17 Sep 2008 23:53:36 -0700 (PDT), sui <go*********@gmail.comwrote:

On Sep 17, 8:04 pm, Peter Pearson <[e-mail]:

>On Wed, 17 Sep 2008 05:28:05 -0700 (PDT), sui <[e-mail]:

[snip]

socket.error: (110, 'Connection timed out')

[snip]

>As a simple connectivity test, you might see whether you can connect
using telnet:

[snip]

even i couldnt connect using telnet....
msg comes host is down

Then your problem is a networking problem. I know even less about
networking than I know about Python. Can you ping the destination?
Perhaps tracepath or traceroute will help you find where your
messages are being refused. Perhaps you are trying to work through
an internet access provider (e.g., ISP) that doesn't allow direct
connections to remote mail servers.

If I might add a grouchy comment, you really should learn a
little about netiquette. When you don't bother to trim the
quoted context or even to punctuate your text, you broadcast
a conspicuous implication that you value your own time much
more than you value the time of the people whose help you're
soliciting, which is incongruous and insulting. A more
carefully presented request might have gotten a response
from someone more knowledgeable -- and less grouchy -- than
me.

--
To email me, substitute nowhere->spamcop, invalid->net.