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How to take a screenshot when except happens and send it in a email? (Python)

I want to take a screenshot using python/selenium when an except happens and send it through an email? How do I do this? I don't want to save it, but I don't mind saving it to a folder but I want to send an email with where the error happened. All examples so far I've seen seem to be saving it to a folder. My code below.
def myfunction(self):
try:
#some code
except NoSuchElementException:
print("error appeared")
#take screenshot here
subject = '')
message = (f"""Here is the screeenshot of the error """)
self.sendEmailBOTS(self.errorReportemail, self.errorReportemailpasswd, self.errorReportemail, subject, message)
for mail in self.ccBots:
self.sendEmailBOTS(self.errorReportemail, self.errorReportemailpasswd, mail, subject, message)
def sendEmailBOTS(self, email, password, send_to, subject, message):
print("Sending email to", send_to)
logging.info("Sending email to")
msg = MIMEMultipart()
msg["From"] = email
msg["To"] = send_to
msg["Subject"] = subject
msg.attach(MIMEText(message, 'plain'))
server = smtplib.SMTP("smtp.gmail.com", 587)
server.starttls()
server.login(email, password)
text = msg.as_string()
server.sendmail(email, send_to, text)
server.quit()

You can try to take a screen shot like this :
myfunction(self):
try:
# some code
except NoSuchElementException:
print("error appeared")
# take screenshot here
driver.save_screenshot('file name here')

def somefunction(self):
imagepath = os.path.join(root, 'Errorimages\\ERROR')
image_screenshot = imagepath + ".png"
self.driver.save_screenshot(image_screenshot)
subject = '')
message = (f"""Here is the screeenshot of the error """)
self.sendEmailBOTS(self.errorReportemail, self.errorReportemailpasswd, self.errorReportemail, subject, message)
for mail in self.ccBots:
self.sendEmailBOTS(self.errorReportemail, self.errorReportemailpasswd, mail, subject, message)
def filefunction(self):
def sendEmailBOTS(self, email, password, send_to, subject, message):
print("Sending email to", send_to)
logging.info("Sending email to")
msg = MIMEMultipart()
msg["From"] = email
msg["To"] = send_to
msg["Subject"] = subject
msg.attach(MIMEText(message, 'plain'))
filename = "ERROR.png"
locationImages = os.path.join(root, 'Errorimages\\ERROR.png')
attachment = open(locationImages, "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("smtp.gmail.com", 587)
server.starttls()
server.login(email, password)
text = msg.as_string()
server.sendmail(email, send_to, text)
server.quit()
This is what I came up with. Its not perfect but it takes the screenshot ,saves it in a folder and then the email sends that picture.

Python Email Notification Not Sending

This code is supposed to send an email to a specified address and when I hard code the "TEXT" & "SUBJECT" it seems to send fine but when I create it as a function and call it it never sends the email and never prints the "Notification Sent" message. What am I missing?
Tried hard coding the TEXT and SUBJECT and it sends fine! NOTE: YOU MUST ENABLE LESS SECURE APPS WHEN USING GMAIL!
import smtplib
class email_thing:
def email_notification(self,SUBJECT,TEXT):
TO = 'email#example.com'
self.SUBJECT = SUBJECT
self.TEXT = TEXT
gmail_sender = 'email#example.com'
gmail_passwd = 'examplepassword'
server = smtplib.SMTP('smtp.gmail.com', 587)
server.ehlo()
server.starttls()
server.login(gmail_sender, gmail_passwd)
return self.SUBJECT
return self.TEXT
BODY = '\r\n'.join(['To: %s' % TO,
'From: %s' % gmail_sender,
'Subject: %s' % SUBJECT,
'',TEXT])
try:
server.sendmail(gmail_sender, [TO], BODY)
print ('Notification Sent!')
except:
print ('error sending mail')
server.quit()
test_send = email_thing()
test_send.email_notification(SUBJECT ='Test Email',TEXT = 'This is a test from python!')

Remove
return self.SUBJECT
return self.TEXT
return exits method at once so code after return is never executed.

Why email is not being send to CC list?

I have the following code where the Cc email list is not working meaning emails are not received to the folks/groups in Cc list,email to To list is working,I just can't figure out what is wrong?any guidance on how to debug and fix it?
import time,smtplib
import pprint,logging,os,time,json
from subprocess import Popen, PIPE, call
from pprint import pprint
from email.mime.text import MIMEText
from email.MIMEMultipart import MIMEMultipart
from email.MIMEBase import MIMEBase
email = 'username2#company.com'
gerriturl = ""
def sendEmail2(type,data):
global originalchange
global gerriturl,email,username1
body = '''%s''' % (data)
#msg = MIMEMultipart(body)
msg = MIMEMultipart()
sender = 'techci#company.com'
receivers = []
cc = ['username1#company.com']
REPLY_TO_ADDRESS = 'embedded-tech-integrators#group.company.com'
if type =='failure':
print("Inside failure email %s"%email)
b = '\U0001F6A8'
print("b.decode('unicode-escape')")
receivers.append('username2#company.com')
#receivers.append(email)
print('%s'%receivers)
msg['Subject'] = '%s AUTO FAILED FOR GERRIT %s :PLEASE TAKE IMMEDIATE ACTION!!!'%(b.decode('unicode-escape'),gerriturl)
msg['From'] = sender
msg['To'] = ', '.join(receivers)
msg['Cc'] = ', '.join(cc)
msg["Content-Type"] = "text/html"
try:
mail = smtplib.SMTP('relay.company.com', 25)
msg.attach(MIMEText(body, 'html'))
msg.add_header('reply-to', REPLY_TO_ADDRESS)
print('Email sent successfully %s %s'%(receivers,cc))
except Exception as e:
logger.error('Problem sending email')
logger.error('%s' % e)
def main():
data = "THIS IS A TEST EMAIL"
sendEmail2('failure',data)
if __name__ == "__main__":
main()

The important thing is to pass the complete list of recipient email addresses to the sendmail() method - you need to concatenate the email addresses for the To, CC and BCC fields into a single list, not a string.
import smtplib
def send_email(subject, body_text, emails_to, emails_cc=[], emails_bcc=[]):
# Configuration
host = "localhost"
port = 1025
from_addr = "nobody#somewhere.com"
# Build the email body
body = "\r\n".join([
"From: %s" % from_addr,
"To: %s" % ', '.join(emails_to),
"CC: %s" % ', '.join(emails_cc),
"Subject: %s" % subject ,
"",
body_text
])
# Prepare the complete list of recipient emails for the message envelope
emails = emails_to + emails_cc + emails_bcc
# Connect to the server
server = smtplib.SMTP(host, port)
# Send the email
server.sendmail(from_addr, emails, body)
# Close the server connection
server.quit()

Populate other email fields like Subject in Python mail function

I have tested this piece of code and it sends the email successfully, it tends to leave the subject fields, cc and bcc fields empty as seen in the photo.
import smtplib
gmail_user = 'dummy#gmail.com'
gmail_password = 'password'
sent_from = 'dummy#gmail.com'
to = ['receiver#gmail.com']
subject = 'OMG Super Important Message'
body = "Hey, what's up?\n\n- You"
email_text = """\
From: %s
To: %s
Subject: %s
%s
""" % (sent_from, ", ".join(to), subject, body)
try:
server = smtplib.SMTP_SSL('smtp.gmail.com', 465)
print("establish connection successful")
server.ehlo()
print("identification successful")
server.login(gmail_user, gmail_password)
print("login successful")
server.sendmail(sent_from, to, email_text)
print("send mail successful")
server.close()
print('Email sent!')
except:
print('Something went wrong...')
Does anyone know how I can fill them up through this script?

For the email subject - there is a specific format to the input arg you provide to server.sendmail that should work. Could you try:
subject = 'This is the email subject'
text = 'This is the email body'
message = "Subject: {}\n\n{}".format(subject, text)
server.sendmail(sender, recipient, message)

If you try to open the source of an email you will see something like this:
Received: from ************.net ([***.**.2.17]) by
*****.****.net ([**.***.224.162]) with mapi id **.03.****.***;
Mon, 22 May 2017 09:14:59 +0200
From: *FROMEMAIL* <******#***.com>
To: *TOEMAIL* <********#***.com>
CC: *CCEMAIL* <********#*****.com>
Subject: E-mail - 150633**0686_****.pdf
...
That is the header of the email, so if you try something like this:
import smtplib
gmail_user = 'dummy#gmail.com'
gmail_password = 'password'
sent_from = 'dummy#gmail.com'
to = ['receiver#gmail.com']
subject = 'OMG Super Important Message'
body = "Hey, what's up?\n\n- You"
cc = "****#***.com"
bcc = "******#*****.com"
email_text = """\
From: %s
To: %s
CC: %s
BCC: %s
Subject: %s
%s
""" % (sent_from, ", ".join(to), cc, bcc,subject, body)
try:
server = smtplib.SMTP_SSL('smtp.gmail.com', 465)
print("establish connection successful")
server.ehlo()
print("identification successful")
server.login(gmail_user, gmail_password)
print("login successful")
server.sendmail(sent_from, to, email_text)
print("send mail successful")
server.close()
print('Email sent!')
except:
print('Something went wrong...')
I think it will work

In python's smtplib why isn't the subject or From field set?

can anyone see why the following code sends emails successfully but it shows up as being from the email address of the sender instead of the name of the sender in the recipients inbox, and the subject shows up as "No Subject" in their inbox. Thanks in advance.
def send_email(destination, subject, message):
from_who = "My Name"
to_send = """\From: %s\nTo: %s\nSubject: %s\n\n%s""" % (from_who, destination, subject, message)
try:
server = smtplib.SMTP("smtp.gmail.com", 587)
server.ehlo()
server.starttls()
server.login(my_email_addr, my_pwrd)
server.sendmail(from_who, destination, message)
server.close()
except:
print "failed"

If you want to use SMTP in python, you need to send the data as a dictionary.
from email.parser import Parser
# This will parse out your to-, from-, and subject fields automatically
headers = Parser().parsestr(to_send)
# This will take care of the to- and from-fields for you
server.send_message(headers)
test.py
from email.parser import Parser
from_whom = "hello#example.com"
destination = "world#example.com"
subject = "Foobar!"
message = "Bar bar binks"
to_send = """From: %s\nTo: %s\nSubject: %s\n\n%s""" % (from_whom, destination, subject, message)
print(to_send)
headers = Parser().parsestr(to_send)
print(headers["To"]) # world#example.com
print(headers["From"]) # hello#example.com
print(headers["Subject"]) # Foobar!
EDIT
Alternatively, you could do this:
to_send = string.join((
"From: %s" % from_whom,
"To: %s" % destination,
"Subject: %s" % subject,
"",
message
), "\r\n")
# ...
server.sendmail(from_whom, [destination], to_send)
# ...
I think the other way is cleaner, but this is up to you.