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python email multiple receiver, multiple attachment for error

def mail():
import os
import pandas as pd
import smtplib
from email.mime.base import MIMEBase
from email.mime.text import MIMEText
from email.utils import formatdate
from email.mime.multipart import MIMEMultipart
from email import encoders
from PyQt5.QtCore import QDate, Qt
path = 'C:/Users/user/Desktop/pdf/'
contact = 'con1.xlsx'
df = pd.read_excel(str(path)+contact, endcoding ='utf8')
df.set_index('unyong', inplace = True)
now = QDate.currentDate()
filenm = [f for f in os.listdir(path) if f.endswith('.pdf')]
unyong_nm = []
for w in filenm:
a = w.find('_')
b = w.rfind('_')
unyong_nm.append(w[a+1:b])
unyong_nm = list(set(unyong_nm))
for i in range(0,len(unyong_nm)):
send_from = 'ss#ddd'
recipients = df.loc[unyong_nm[i],'email']
send_to = ",".join(recipients)
attach = [s for s in filenm if s.find(unyong_nm[i]) >-1 ]
username = 'sss#ssss'
password = 'sss'
subject = ('111'+now.toString('yyyy.MM')+'_'+unyong_nm[i]+str(i+1))
text = ('hi')
msg = MIMEMultipart()
msg['From'] = send_from
msg['To']= send_to
msg['Subject'] = subject
msg['Date']=formatdate(localtime=True)
filename_match = [s for s in filenm if s.find(unyong_nm[i]) >-1 ]
for file in filename_match:
part =MIMEBase('application','octet-stream')
part.set_payload(open(str(path)+file, 'rb').read())
encoders.encode_base64(part)
part.add_header('Content-Disposition','attachment', filename =file)
msg.attach(part)
msg.attach(MIMEText(text))
mailServer = smtplib.SMTP("smtp.sssss.com", 587)
mailServer.ehlo()
mailServer.starttls()
mailServer.ehlo()
mailServer.login(username,password)
mailServer.sendmail(send_from, send_to, msg.as_string())
mailServer.close()
hi i have a problem with the email with attachment for statement.
the results of below def mail(),
multiple emails was sent to one person. (<- this is a error)
I want to send a each specific reciever with specific multiple attachments only once
why multiple email sented with diffrent number of attachements to one person.
the reciever have the 2 emails with a 1 attachment and, simultaneouly 2 attachment.
I want to send a email containg 2 attachments
please help me.
*CF) path containg thoes files:
['2221_sss_love.pdf', '2221_sss_happy.pdf', '2221_ddd_sad.pdf', '2221_ddd_lucky.pdf', 'con1.xlsx']
*result
unyong_nm = ['sss','ddd']
filenm = ['2221_sss_love.pdf', '2221_sss_happy.pdf', '2221_ddd_sad.pdf', '2221_ddd_lucky.pdf']
*CF) con1.xlsx file contenxt:
unyong email
sss 111#aaa
sss 777#bbb
ddd 666#sss
ddd 444#ccc

The code is sending an email for each attachment. By dedenting the mail-sending code, an email will be sent for each set of attachments grouped by 'ddd' or 'sss'.
for file in filename_match:
part = MIMEBase("application", "octet-stream")
part.set_payload(open(file, "rb").read())
encoders.encode_base64(part)
part.add_header("Content-Disposition", "attachment", filename=file)
msg.attach(part)
msg.attach(MIMEText(text))
# Send mail outside the file grouping loop
mailServer = smtplib.SMTP("localhost", 1025)
mailServer.ehlo()
mailServer.sendmail(send_from, send_to, msg.as_string())
mailServer.close()
The recipient selection code
recipients = df.loc[unyong_nm[i],'email']
send_to = ",".join(recipients)
might need to be changed, I can't tell because the contents of the dataframe aren't provided in the question.

How can I send email to multi addresses by smtplib?

I use bellow code I can send email to one account, but how can I send to multi account?
import smtplib
from email.mime.text import MIMEText
_user = "67676767#qq.com" #
_pwd = "orjxywulscbxsdwbaii" #
_to = linux213#126.com" #
msg = MIMEText("content") #
msg["Subject"] = "邮件主题测试" #
msg["From"] = _user
msg["To"] = _to
try:
s = smtplib.SMTP_SSL("smtp.qq.com", 465)
s.login(_user, _pwd)
s.sendmail(_user, _to, msg.as_string())
s.quit()
print ("Success!")
except smtplib.SMTPException as e:
print ("Falied,%s" % e)

Try this:
import smtplib
from email.mime.text import MIMEText
s = smtplib.SMTP('smtp.qq.com')
s.set_debuglevel(1)
msg = MIMEText("""body""")
sender = 'me#example.com'
recipients = ['k.ankur#abc.com', 'a.smith#abc.co.in']
msg['Subject'] = "subject line"
msg['From'] = sender
msg['To'] = ", ".join(recipients)
s.sendmail(sender, recipients, msg.as_string())

BCC not hidden on gmail using python smtplib

I am working on a python script to send email to my customer with a survey. I will send only one email with all my customers's emails in the BCC field so that I do not need to loop through all the emails. Everything works fine when I tested sending emails to my company's coleagues and also when I sent to my personal email, but whenever I send to a gmail account, the BCC field appears to not be hidden and show all the emails. I found this post Email Bcc recipients not hidden using Python smtplib and tried that solution as well, but as I am using a html body email, the emails were shown inside the body. Can anyone help me on this one?
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
def send_survey_mail():
template_path = 'template.html'
background_path = 'image.png'
button_path = 'image2.png'
try:
body = open(template_path, 'r')
msg = MIMEMultipart()
msg['Subject'] = 'Customer Survey'
msg['To'] = ', '.join(['myemail#domain.com.br', 'myemail2#domain.com'])
msg['From'] = 'mycompany#mycompany.com.br'
msg['Bcc'] = 'customer#domain.com'
text = MIMEText(body.read(), 'html')
msg.attach(text)
fp = open(background_path, 'rb')
img = MIMEImage(fp.read())
fp.close()
fp2 = open(button_path, 'rb')
img2 = MIMEImage(fp2.read())
fp2.close()
img.add_header('Content-ID', '<image1>')
msg.attach(img)
img2.add_header('Content-ID', '<image2>')
msg.attach(img2)
s = smtplib.SMTP('smtpserver')
s.sendmail('mycompany#mycompany.com.br',
['myemail#domain.com.br', 'myemail2#domain.com', 'customer#domain.com'],
msg.as_string())
s.quit()
except Exception as ex:
raise ex
send_survey_mail()
I removed the following line from the code and tried again. Now the email is not sent to my customer's Gmail email.
msg['Bcc'] = 'customer#gmail.com'

Just do not mention bcc mails in msg['To'] or msg['Cc']. Do it only in server.sendmail()
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders
from_addr = "your#mail.com"
to_addr = ["to#mail.com", "to2#mail.com"]
msg = MIMEMultipart()
msg['From'] = from_addr
msg['To'] = to_addr
msg['Subject'] = "SUBJECT"
body = "BODY"
msg.attach(MIMEText(body, 'plain'))
filename = "FILE.pdf"
attachment = open('/home/FILE.pdf', "rb")
part = MIMEBase('application', 'octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition', "attachment; filename= %s" % filename)
msg.attach(part)
server = smtplib.SMTP('.....', 587)
server.starttls()
server.login(from_addr, 'yourpass')
text = msg.as_string()
server.sendmail(from_addr, to_addr + [bcc#mail.com], text)
server.quit()

Did you try not to define the msg['BCC'] field? Setting this field forces it to be included. It is sufficient that the BCC email address is in the sendmail command's destination address list. Take a look at this question.

MAIL_FROM = default#server.com
MAIL_DL = default#server.com
def send(to, cc, bcc, subject, text, html):
message = MIMEMultipart("alternative")
message["Subject"] = subject
message["From"] = MAIL_FROM
message["To"] = to
message["Cc"] = cc + "," + MAIL_DL
if html is not None:
body = MIMEText(html, "html")
else:
body = MIMEText(text)
message.attach(body)
server = smtplib.SMTP(MAIL_SERVER)
server.set_debuglevel(1)
server.sendmail(MAIL_DL, to.split(",") + bcc.split(","), message.as_string())
server.quit()
return {
"to": to,
"cc": cc,
"bcc": bcc,
"subject": subject,
"text": text,
"html": html,
}

send mail using python

Problem: when i send mail to user then from user name not seen in to user inbox only show email-id but i need user name of sender
from: demo#gmail.com username: Demo
To: demotest#gmail.com
CODE
import smtplib
fromaddr = From
toaddrs = To
msg = 'Why,Oh why!'
username = From
password = *******
server = smtplib.SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()
server.login(username, password)
server.sendmail(fromaddr, toaddrs, msg)
server.quit()

smtplib does not include automatically any header, and you need to include a From: header, so you have to put one by yourself doing something like:
# Add the From: and To: headers at the start!
msg = ("From: %s\r\nTo: %s\r\n\r\n"
% (fromaddr, ", ".join(toaddrs)))
As you can se in the DOCS.

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
fromaddr = 'demo#gmail.com'
toaddrs = 'demotest#gmail.com'
msg = MIMEMultipart('alternative')
msg['Subject'] = "Link"
msg['From'] = "good morning" #like name
msg['To'] = "GGGGGG"
body = MIMEText("example email body")
msg.attach(body)
username = 'demo#gmail.com'
password = ''
server = smtplib.SMTP_SSL('smtp.googlemail.com', 465)
server.login(username, password)
server.sendmail(fromaddr, toaddrs, msg.as_string())
server.quit()

You just need to create the message correctly. I think the most convenient way to do it is using a special object for message. I placed a class which perhaps may help you to send messages in your project.
import os
import smtplib
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart
class EmailSender(object):
def __init__(self, subject, to, config):
self.__subject = subject
self.__to = tuple(to) if hasattr(to, '__iter__') else (to,)
self.__from = config['user']
self.__password = config['password']
self.__server = config['server']
self.__port = config['port']
self.__message = MIMEMultipart()
self.__message['Subject'] = self.__subject
self.__message['From'] = self.__from
self.__message['To'] = ', '.join(self.__to)
def add_text(self, text):
self.__message.attach(
MIMEText(text)
)
def add_image(self, img_path, name=None):
if name is None:
name = os.path.basename(img_path)
with open(img_path, 'rb') as f:
img_data = f.read()
image = MIMEImage(img_data, name=name)
self.__message.attach(image)
def send(self):
server = smtplib.SMTP_SSL(self.__server, self.__port)
server.login(self.__from, self.__password)
server.sendmail(self.__from, self.__to, self.__message.as_string())
server.close()
sender = EmailSender("My letter", "my_target#email", {
'user': "from#email",
'password': "123456",
'server': "mail.google.com"
'port': 465
})
sender.add_text("Why,Oh why!")
sender.send()

Or go the easy way, by install yagmail and
Given:
To = 'someone#gmail.com'
From = 'me#gmail.com'
pwd = '******'
alias = 'someone'
Run:
import yagmail
yag = yagmail.SMTP(From, pwd)
yag.send({To: alias}, 'subject', 'Why,Oh why!')
Install might be done by pip install yagmail

send email to recipient from .txt in python 3.3

I am trying to send an email with smtp in python 3.3 to a recipient listed in a text file.
The error I receive is:
session.sendmail(sender, recipient, msg.as_string())
smtplib.SMTPRecipientsRefused: {}
Where is the error in sendmail? Thanks!
Full code below:
#!/usr/bin/python
import os, re
import sys
import smtplib
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
mails = open('/path/emails.txt','r+')
mailList = mails.read()
mailList = [i.strip() for i in mails.readlines()]
directory = "/path/Desktop/"
SMTP_SERVER = 'smtp.gmail.com'
SMTP_PORT = 587
sender = 'Sender#gmail.com'
password = "Sender'sPassword"
recipient = [i.strip() for i in mails.readlines()]
subject = 'Python (-++-) Test'
message = 'Images attached.'
def main():
msg = MIMEMultipart()
msg['Subject'] = 'Python (-++-) Test'
msg['To'] = recipient
msg['From'] = sender
files = os.listdir(directory)
pngsearch = re.compile(".png", re.IGNORECASE)
files = filter(pngsearch.search, files)
for filename in files:
path = os.path.join(directory, filename)
if not os.path.isfile(path):
continue
img = MIMEImage(open(path, 'rb').read(), _subtype="png")
img.add_header('Content-Disposition', 'attachment', filename=filename)
msg.attach(img)
part = MIMEText('text', "plain")
part.set_payload(message)
msg.attach(part)
session = smtplib.SMTP(SMTP_SERVER, SMTP_PORT)
session.ehlo()
session.starttls()
session.ehlo
session.login(sender, password)
session.sendmail(sender, recipient, msg.as_string())
session.quit()
if __name__ == '__main__':
main()

You want to double-check your recipients code. It looks like you're trying to consume the contents of the file more than once, which won't work--the file objects should be understood as a stream, not a block of data, so once you've done f.read() or [i.strip() for i in mails.readlines()] one time, that stream is empty, so doing it a second time is going to produce an empty list. You should check this yourself by printing recipient
then try this:
mails = open('/path/emails.txt','r+')
#mailList = mails.read()
#mailList = [i.strip() for i in mails.readlines()]
directory = "/path/Desktop/"
SMTP_SERVER = 'smtp.gmail.com'
SMTP_PORT = 587
sender = 'Sender#gmail.com'
password = "Sender'sPassword"
recipient = [i.strip() for i in mails.readlines()]
print(recipient)
subject = 'Python (-++-) Test'
message = 'Images attached.'
Now you should have the a populated recipient list, and on to the next issue!