On Mar 20, 12:50 pm, "Boudreau, Emile" <[EMAIL PROTECTED]>
wrote:
> Thanks for the reply. When I use the instruction from that list this is
> the email I receive. I'm using Outlook.
>
> [EMAIL PROTECTED]
> To:
> ------------------------------
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> Subject: Hello!
>
> This Still DOESN't Work!!!!!!!!!!!!
>
> It's just adding the "From" "To" "Subject" in the message itself. I want
> to have each field at the correct place and then just the msg in the
> body.
>
> Any Help?? Thanks
>
> Emile Boudreau
>
> -----Original Message-----
> From: [EMAIL PROTECTED]
>
> [mailto:[EMAIL PROTECTED] On
> Behalf Of Fredrik Lundh
> Sent: Tuesday, March 20, 2007 11:08 AM
> To: [EMAIL PROTECTED]
> Subject: Re: Subject line with smtplib.sendmail()
>
> Boudreau, Emile wrote:
>
> > I'm trying to send mail from my server to my machine with test
> > results. I can send the email no problem however, the email doesn't
> > contain a "recipient list" or a "subject line". I was wondering how
> > would I go about getting the information on the actual "To" and
> > "Subject" lines so that I know to whom the email was sent and the
> > subject line without opening the email?
>
> you have to add the headers yourself. see the example in the library
> reference, or this FAQ entry:
>
> http://effbot.org/pyfaq/how-do-i-send-mail-from-a-python-script
>
> </F>
>
> --http://mail.python.org/mailman/listinfo/python-list
>
> This message may contain privileged and/or confidential information. If
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> may not use, copy, disseminate or distribute it; do not open any attachments,
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I'm not sure what the problem is. But here's how we do it at my place
of work:
import smtplib
import string
def sendMail(subject, body, TO = "[EMAIL PROTECTED]",
FROM="[EMAIL PROTECTED]"):
HOST = "mailserver"
BODY = string.join((
"From: %s" % FROM,
"To: %s" % TO,
"Subject: %s" % subject,
"",
body
), "\r\n")
server = smtplib.SMTP(HOST)
server.sendmail(FROM, [TO], BODY)
server.quit()
This works well for us.
Mike
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