so I'm working on something that uses regex to search something from an email, which is fetched via imaplib module. Right now I can't get it to work, even after using str() function.
result, data = mail.fetch(x, '(RFC822)')
eemail = email.message_from_bytes(data[0][1])
print(str(eemail))
trying to regex it:
print(re.search("button", eemail))
Regex gives me no matches even after making the email a string object.
This is what I use:
import imaplib
import email
import re
mail = imaplib.IMAP4_SSL(SMTP_SERVER, SMTP_PORT)
mail.login(FROM_EMAIL,FROM_PWD)
mail.select('inbox')
status, response = mail.search(None, '(UNSEEN)')
unread_msg_nums = response[0].split()
for e_id in unread_msg_nums:
_, response = mail.fetch(e_id, '(UID BODY[TEXT])')
b = email.message_from_string(response[0][1])
if b.is_multipart():
for payload in b.get_payload(decode=True):
print(re.search("button", payload.get_payload(decode=True)))
else:
print(re.search("button", b.get_payload(decode=True)))
Related
Windows. Python 3.9.
As a value of mail subject I get other characters instead of Polish characters - I get:
Odpowied�� automatyczna: "Re: Program licz��cy ceny i sprzeda�� w allegro dla EAN��w"
instead of:
Odpowiedź automatyczna: "Re: Program liczący ceny i sprzedaż w allegro dla EANów"
How to make it correct? Should I apply some codepage information somewhere?
I notice all out dictionary values are string except for the subject which is of type Header.
import imaplib, email
mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login('user', 'pwd')
mail.select('inbox')
data = mail.search(None, 'ALL')
_, data = mail.fetch(str(7), '(RFC822)')
message = email.message_from_bytes(data[0][1])
out = {
'from': message['from'],
'subject': message['subject'],
'to': message['Delivered-To'],
'datetime': message['Date'],
'cc': message['Cc']
}
if understand correctly you need to decode bytes.
try something like
from email.header import decode_header
subject, encoding = decode_header(message["subject"])[0]
if isinstance(subject, bytes):
subject = subject.decode(encoding)
I get this error when sending emails via smptplib in Python to a certain list
SMTPRecipientsRefused: {'': (421, b'4.7.0 Too many protocol errors (6) on this connection, closing transmission channel.')}?
I'm using Office365 SMTP details and a snippet of the code is below:-
import smtplib, ssl
from email.message import EmailMessage
import getpass
ids = df['IDs']
emails_to = df['Emails']
namesofcompanies = df["CompanyNames"]
sendfrom = df["SenderList"]
date_7days = (datetime.now() + timedelta(days=7)).strftime('%d/%m/%Y')
date_14days = (datetime.now() + timedelta(days=13)).strftime('%d/%m/%Y')
email_pass = input() #Office 365 password
context=ssl.create_default_context()
for i in range(len(emails_to)): # iterate through the records
# for every record get the name and the email addresses
ID = str(ids[i])
Emaitstosendto = emails_to[i]
companynames = namesofcompanies[i]
tosendfrom = sendfrom[i]
if my_files_dict.get(ID): #Looks for attachments in the same folder with same name as the corresponding record
smtp_ssl_host = 'smtp.office365.com'
smtp_ssl_port = 587
email_login = "xxx#xxx.com" #Office 365 email
email_from = tosendfrom
email_to = Emaitstosendto
msg = MIMEMultipart()
msg['Subject'] = "Received Emails between "+date_7days+" - "+date_14days
msg['From'] = email_from
msg['To'] = email_to
msg['X-Priority'] = '2'
text = ("XXXX,\n"
f"xxxxxx\n\n")
msg.attach(MIMEText(text))
filename = my_files_dict.get(ID)#Files in the folder matching the ID
fo = open(filename,'rb')
s2 = smtplib.SMTP(smtp_ssl_host, smtp_ssl_port)
s2.starttls(context=context)
s2.login(email_login, email_pass)
attachment = email.mime.application.MIMEApplication(fo.read(),_subtype="xlsx")
fo.close()
attachment.add_header('Content-Disposition','attachment',filename=filename)
msg.attach(attachment)
s2.send_message(msg)
s2.quit()
On average, I will be sending emails to a list separated by a semi-colon(;) of about 8 emails per record. This means that for each attachment, I'll send to about 8 emails and I'm to do it for about 70 of such contacts. In total, that will be about 560 emails. Nothing gets sent out I get the above error the moment I log in. On the contrary, when I try test sending it to a list of 3 emails in the test emails column, the same emails go out very well. Can anyone point to where I may not be getting it right? I suspect either the list of emails is too long or is something with the email addresses thus the protocol error? Is this an SMTPlib limitation?
What you specify in MIMEMultipart() is what appears in the message header, but this is not always equal to the list of recipients. You can try to change server.send_message(msg) to server.sendmail(sender,recipients,msg.as_string())
Keep in mind that sendmail() requires a list of recipients, whereas msg['To'] should be set as one string, so if your variable email_to is comma separated, you should write it like:
s2.sendmail(email_from,email_to.split(','),msg.as_string())
More details about sendmail() can be found here.
I resolved this error by creating a list and joined all addresses from a tuple into a string, using a comma character as separator.
family = the list of all recipients' email addresses
family = [
'name1#example.com',
'name2#example.com',
'name3#example.com',
'name4#example.com',
'name5#example.com',
'name6#example.com',
'name7#example.com',
'name8#example.com'
]
msg['To'] =', '.join(family)
How can I see what labels an email has?
mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login('myaccountxyz#gmail.com', mypassword)
mail.select("my-folder") # finds emails with this label
result, data = mail.uid('search', None, 'all')
for email_uid in data[0].split():
result, data_single = mail.uid('fetch', email_uid, '(RFC822)')
raw_email = data_single[0][1]
email_message = email.message_from_string(raw_email)
sender = email_message['From']
# get list of this email's labels
I haven't tried this code myself, but according to Google IMAP Extensions, you should be able to just fetch the X-GM-LABELS item:
typ, dat = mail.uid('fetch', email_uid, 'X-GM-LABELS')
I am using python standard email parsing library to parse the raw email that I am getting from amazon ses mail service.
Below is my code for the same.
import json
import email
from email.Utils import parseaddr
def parse(raw_email):
message = email.message_from_string(raw_email)
text_plain = None
text_html = None
for part in message.walk():
if part.get_content_type() == 'text/plain' and text_plain is None:
text_plain = part.get_payload()
if part.get_content_type() == 'text/html' and text_html is None:
text_html = part.get_payload()
parsed_email_object = {
'to': parseaddr(message.get('To'))[1],
'from': parseaddr(message.get('From'))[1],
'delivered to': parseaddr(message.get('Delivered-To'))[1],
'subject': message.get('Subject'),
'text_plain': text_plain,
'text_html': text_html,
}
json_string = json.dumps(parsed_email_object)
return json_string
when I am parsing my raw email, it is not parsing 100%, it is giving me unwanted characters like this
this is a replyo from the gmail indbo asdf asdf asdfa sdfa=
sd sdfa sdfa fasd
=C2=A0dfa sf asdf
a sdfas
<= div>f asdf=C2=A0
Is there anything else like some decoding option to parse it correctly.
Making my comment as an answer so that it gets noticed.
part.get_payload(decode=True).decode(part.get_content_charset())
This will solve the issue of encoding
I am using SendGrid for Python. I want to CC some people in an email. It seems like they may no longer support CC'ing on emails, though I'm not positive if that's true? But surely there is a work around to it somehow, but I am surprised I can't find much support on this.
Here is my basic code:
sg = sendgrid.SendGridAPIClient(apikey='*****')
from_email = Email(sender_address, sender_name)
to_email = Email(email_address)
subject = subject
content = Content("text/plain", email_message)
mail = Mail(from_email, subject, to_email, content)
response = sg.client.mail.send.post(request_body=mail.get())
How can I modify this so it will CC someone on an email?
Using the SendGrid's Personalization() or Email() class did not work for me. This is how I got it to work:
from sendgrid import SendGridAPIClient
from sendgrid.helpers.mail import Mail, Cc
# using a list of tuples for emails
# e.g. [('email1#example.com', 'email1#example.com'),('email2#example.com', 'email2#example.com')]
to_emails = []
for r in recipients:
to_emails.append((r, r))
# note the Cc class
cc_emails = []
for c in cc:
cc_emails.append(Cc(c, c))
message = Mail(
from_email=from_email,
to_emails=to_emails,
subject='My Subject',
html_content=f'<div>My HTML Email...</div>'
)
if cc_emails:
message.add_cc(cc_emails)
try:
sg = SendGridAPIClient(os.getenv('SENDGRID_API_KEY'))
sg.send(message)
except Exception as e:
print(f'{e}')
Hopefully this helps someone.
I resolved it. Santiago's answer got me mostly there, but here is what I needed to do:
sg = sendgrid.SendGridAPIClient(apikey='****')
from_email = Email(sender_address, sender_name)
to_email = Email(to_email)
cc_email = Email(cc_email)
p = Personalization()
p.add_to(to_email)
p.add_cc(cc_email)
subject = subject
content = Content("text/plain", email_message)
mail = Mail(from_email, subject, to_email, content)
mail.add_personalization(p)
response = sg.client.mail.send.post(request_body=mail.get())
If you don't include the p.add_to(to_email) it rejects it because there is no "to email" in the personalization object. Also, if you don't include the "to_email" inside the mail object it rejects it because it is looking for that argument, so you have to be a bit redundant and define it twice.
I've been looking at the code: https://github.com/sendgrid/sendgrid-python/blob/master/examples/mail/mail.py
And it looks like you can do that by adding a personalization to the mail, for example:
cc_email = Email(cc_address)
p = Personalization()
p.add_cc(cc_email)
mail.add_personalization(p)
Based on the answers here you can CC to email if you add another email to 'to_email'.
If you want to cc multiple user then in djanogo using sendgrid you need to import the below line
the function that will be used to send the mail
and finally how you ned to send the data paramters to the above function so that it can CC the person
email = send_sandgridmail(sender=sender,receiver=receivers,subject=subject,content=message,reply_to=sender,cc=[admin_mail_account_mail,"rawatanup918#gmail.com"],attachment=None)
i hope this'll help.simplified from #anurag image script
import os
from sendgrid import SendGridAPIClient
from sendgrid.helpers.mail import To,Mail,ReplyTo,Email,Cc
def send_sandgridmail (sender, receiver, subject, content, reply_to=None, cc=[], attachment=None) :
# content = convert_safe_text(content)
# to email = To(receiver)
message = Mail(
from_email=str(sender),
to_emails=receiver,
subject= str(subject),
html_content = content)
if reply_to:
message.reply_to= ReplyTo(reply_to)
if attachment:
message.add_attachment (attachment)
if len(cc):
cc_mail = []
for cc_person in cc:
cc_mail.append(Cc(cc_person, cc_person))
message.add_cc (cc_mail)
try:
SENDGRID_API_KEY = 'your sendgrid api key'
sg= SendGridAPIClient (SENDGRID_API_KEY)
response= sg.send(message)
print (response.status_code)
# print (response.body)
# print (response.headers)
except Exception as e:
print(e)
return response