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I am trying to make a email chatbot but it spams how could i fix this?

I am trying to build a email chatbot but it has this bug where after it sends the first message, and then gets a response it keeps spamming the answered to the response it got until it gets another response which then it repeats again I was thinking to solve this I should use a variable which detects emails and later down the code a condition that responds only if a email is received, does anyone have any idea on how I could fix this? Thanks
def receive_email():
try:
mail = imaplib.IMAP4_SSL("smtp.gmail.com")
mail.login(email_address, email_password)
mail.select('inbox')
#searches inbox
status, data = mail.search(None, 'Recent')
mail_ids = data[0].split()
latest_email_id = mail_ids[-1]
status, data = mail.fetch(latest_email_id, '(RFC822)')
#gets message
for response_part in data:
if isinstance(response_part, tuple):
msg = email.message_from_bytes(response_part[1])
sender = msg['from']
subject = msg['subject']
if msg.is_multipart():
for part in msg.get_payload():
if part.get_content_type() == 'text/plain':
return part.get_payload()
message = msg.get_payload()
return message,
except Exception as e:
print("Error: ", e)
print("Could not receive email")
return None, None

This is the usual problem for an email autoresponder, if I understand you correctly, and RFC 3834 offers good advice.
Since answers should be self-contained I offer a summary:
Add the Auto-Submitted: auto-replied header field on your outgoing messages. Any value other than no will prevent well-written autoresponders from replying to your outgoing messages.
Set the \answered flag on the message you reply to, immediately before you send the reply.
Change the search key from recent to unanswered not header "auto-submitted" "". unanswered means that the search won't match the messages on which you set the \answered flag, not header "auto-submitted" "" means that you'll not match messages that contain any auto-submitted header field.
Direct your replies to the address in return-path or sender, not the one in from. This is a matter of convention. Auto-submitted mail will often have a special return-path that points to an address that never sends any autoreply.
You may also extend the search key with more details from RFC 3834. The one I suggest should work, but not header "precedence" "junk" will for example prevent your code from replying to a bit of autogenerated mail. Sendgrid and its friends also add header fields you may want to look for and exclude.
If the incoming message has headers like this (use the "view headers" function of most mail readers to see it):
From: example#example.com
Subject: Weekend
To: srtai22#gmail.com
Message-id: <56451182ae7a62978cd6f6ff06dd21e0#example.com>
Then your reply should have headers like this:
Return-Path: <>
From: srtai22#gmail.com
To: example#example.com
Auto-Submitted: auto-replied
Subject: Auto: Weekend
References: <56451182ae7a62978cd6f6ff06dd21e0#example.com>
There'll be many more fields in both, of course. Your reply's return-path says that nothing should respond automatically, From and To are as expected, auto-submitted specifies what sort of response this is, subject doesn't matter very much but this one's polite and well-behaved, and finally references links to the original message.

Transfer an email with Python

I've tried with no conclusions to resend emails with Python.
Once I've logged in SMTP and IMAP with TLS, this is what I have written:
status, data = self._imapserver.fetch(id, "(RFC822)")
email_data = data[0][1]
# create a Message instance from the email data
message = email.message_from_string(email_data)
# replace headers (could do other processing here)
message.replace_header("From", 'blablabla#bliblibli.com')
message.replace_header("To", 'blobloblo#blublublu.com')
self._smtpserver.sendmail('blablabla#bliblibli.com', 'blobloblo#blublublu.com', message.as_string())
But the problem is that the variable data doesn't catch the information from the email, even if the ID is the one I need.
It tells me:
b'The specified message set is invalid.'
How can I transfer an email with Python?

Like the error message says, whatever you have in id is invalid. We don't know what you put there, so all we can tell you is what's already in the error message.
(Also, probably don't use id as a variable name, as you will shadow the built-in function with the same name.)
There are additional bugs further on in your code; you need to use message_from_bytes if you want to parse it, though there is really no need to replace the headers just to resend it.
status, data = self._imapserver.fetch(correct_id, "(RFC822)")
self._smtpserver.sendmail('blablabla#bliblibli.com', 'blobloblo#blublublu.com', data[0][1])
If you want to parse the message, you should perhaps add a policy argument; this selects the modern EmailMessage API which was introduced in Python 3.6.
from email.policy import default
...
message = email.message_from_bytes(data[0][1], policy=default)
message["From"] = "blablabla#bliblibli.com"
message["To"] = "blobloblo#blublublu.com"
self._smtpserver.send_message(message)
The send_message method is an addition to the new API. If the message could contain other recipient headers like Cc:, Bcc: etc, perhaps using the good old sendmail method would be better, as it ignores the message's headers entirely.

How to read in whole message not just snippet

msg = service.users().messages().get(userId='me', id=message['id']).execute()
print(msg['snippet'])
I am currently using the above code, which doesn't get the whole message. I have seen in the documentation that the google API has raw and full options, but the raw option doesn't print in a readable way and I cannot get the full option to work.
Thank you !

This is how worked for me:
# Gets message header first
msg_header = service.users().messages().get(
userId=user_id,
id=msg_id,
format="full",
metadataHeaders=None
).execute()
# Gets message body from header
body = base64.urlsafe_b64decode(msg_header.get("payload").get("body")\
.get("data").encode("ASCII")).decode("utf-8")
The body comes in HTML so, in my case, I use BeautifulSoup to extract the information I need, like below:
soup = bs(body, 'html.parser')
# Loop on e-mail table
for row in soup.findAll('tr'):
aux = row.findAll('td')
info[aux[0].string] = aux[1].string
The information extraction will depend on the pattern of the message. In my case, all messages that I'm getting have the same pattern.

Receive and Print a Twilio SMS with Python

For clarification, I don't want to reply to the SMS. Every tutorial or document I've looked at is about setting up a port to listen on.
What I'm trying to do is just get the SMS and print it. I can send them fine and without problems.
Here is my sending function, and it works.
def send():
message = client.messages \
.create(
body=sendMSG,
from_='MY_TWILIO_NUMBER',
to='MY_PERSONAL_NUMBER'
)
print(message.sid)
How would you receive an SMS without Flask? Is there a way to do something similar to this method below just for receiving?
def receive():
message = client.messages \
.recieve(
from_='MY_PERSONAL_NUMBER',
to='MY_TWILIO_NUMBER'
)
print(message.sid)

I have not personally tried to get SMS messages from the logs before, always getting it directly through a webhook, but from what I see, it appears the command you might be looking for is list(). You can add filters, as shown in the API docs, and there are three filtering options. You can filter by DateSent, To, or From.
I have not tried this, but it would seem that the way to use this would be the following (adjusted from the code they supply):
# Download the helper library from https://www.twilio.com/docs/python/install
from twilio.rest import Client
# Your Account Sid and Auth Token from twilio.com/console
account_sid = 'ACXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX'
auth_token = 'your_auth_token'
client = Client(account_sid, auth_token)
messages = client.messages.list(from='MY_PERSONAL_NUMBER', to='MY_TWILIO_NUMBER')
for record in messages:
print(record.sid)
If that doesn't work, the variables they use are actually capitalized "To" and "From", so you might try that.
After looking at that a bit, you might be looking more for this:
received = client.messages.list(to='MY_TWILIO_NUMBER')
sent = client.messages.list(from='MY_PERSONAL_NUMBER')
That will separate out those sent to you, and those sent from you

How to receive mail using python

I would like to receive email using python. So far I have been able to get the subject but not the body. Here is the code I have been using:
import poplib
from email import parser
pop_conn = poplib.POP3_SSL('pop.gmail.com')
pop_conn.user('myusername')
pop_conn.pass_('mypassword')
#Get messages from server:
messages = [pop_conn.retr(i) for i in range(1, len(pop_conn.list()[1]) + 1)]
# Concat message pieces:
messages = ["\n".join(mssg[1]) for mssg in messages]
#Parse message intom an email object:
messages = [parser.Parser().parsestr(mssg) for mssg in messages]
for message in messages:
print message['subject']
print message['body']
pop_conn.quit()
My issue is that when I run this code it properly returns the Subject but not the body. So if I send an email with the subject "Tester" and the body "This is a test message" it looks like this in IDLE.
>>>>Tester >>>>None
So it appears to be accurately assessing the subject but not the body, I think it is in the parsing method right? The issue is that I don't know enough about these libraries to figure out how to change it so that it returns both a subject and a body.

The object message does not have a body, you will need to parse the multiple parts, like this:
for part in message.walk():
if part.get_content_type():
body = part.get_payload(decode=True)
The walk() function iterates depth-first through the parts of the email, and you are looking for the parts that have a content-type. The content types can be either text/plain or text/html, and sometimes one e-mail can contain both (if the message content_type is set to multipart/alternative).

The email parser returns an email.message.Message object, which does not contain a body key, as you'll see if you run
print message.keys()
What you want is the get_payload() method:
for message in messages:
print message['subject']
print message.get_payload()
pop_conn.quit()
But this gets complicated when it comes to multi-part messages; get_payload() returns a list of parts, each of which is a Message object. You can get a particular part of the multipart message by using get_payload(i), which returns the ith part, raises an IndexError if i is out of range, or raises a TypeError if the message is not multipart.
As Gustavo Costa De Oliveir points out, you can use the walk() method to get the parts in order -- it does a depth-first traversal of the parts and subparts of the message.
There's more about the email.parser module at http://docs.python.org/library/email.message.html#email.message.Message.

it also good return data in correct encoding in message contains some multilingual content
charset = part.get_content_charset()
content = part.get_payload(decode=True)
content = content.decode(charset).encode('utf-8')

Here is how I solved the problem using python 3 new capabilities:
import imaplib
import email
mail = imaplib.IMAP4_SSL('imap.gmail.com')
mail.login(username, password)
mail.select(readonly=True) # refresh inbox
status, message_ids = mail.search(None, 'ALL') # get all emails
for message_id in message_ids[0].split(): # returns all message ids
# for every id get the actual email
status, message_data = mail.fetch(message_id, '(RFC822)')
actual_message = email.message_from_bytes(message_data[0][1])
# extract the needed fields
email_date = actual_message["Date"]
subject = actual_message["Subject"]
message_body = get_message_body(actual_message)
Now get_message_body is actually pretty tricky due to MIME format. I used the function suggested in this answer.
This particular example works with Gmail, but IMAP is a standard protocol, so it should work for other email providers as well, possibly with minor changes.

if u want to use IMAP4. Use outlook python library, download here : https://github.com/awangga/outlook
to retrieve unread email from your inbox :
import outlook
mail = outlook.Outlook()
mail.login('emailaccount#live.com','yourpassword')
mail.inbox()
print mail.unread()
to retrive email element :
print mail.mailbody()
print mail.mailsubject()
print mail.mailfrom()
print mail.mailto()