We Keep Coding

Can't open attachments using SMTPLIB

These are the imports I used to send the email:
import tkinter as tk
from tkinter import filedialog
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
import email, smtplib, ssl
from email import encoders
import os
I have made a script where you can add attachments to emails using tkfiledialog to select files using this function:
def browse_file(): # Select a file to open
global filelist
filelist = filedialog.askopenfilenames()
files_attached_tk.set("Files Attached: " + str(len(filelist)))
This is the part of the script that attaches and sends the file(s): (For and With are on same indentation)
for file in filelist:
attachment_part = MIMEBase("application", "octet-stream")
attachment_part.set_payload(open(file, "rb").read())
encoders.encode_base64(attachment_part)
attachment_part.add_header("Content-Disposition", "attachment; filename='%s'" % os.path.basename(file))
message.attach(attachment_part)
# Create Server Connection
with smtplib.SMTP_SSL("smtp.gmail.com", 465, context=context) as server:
server.login(config.email_sender, config.email_password)
server.sendmail(
sender_email, reciever_email, message.as_string()
)
The problem is that, the files do send, but, they appear to be wrapped in ' ' in the email attachment. They look like this: 'document.pdf' which makes the document unreadable, for example, it does not say PDF File in the email because it is wrapped in ' '.
I have managed to open the files on my computer but I cannot open them on my phone. How would I be able to remove the ' ' from the file name? I have tried to do os.path.basename(file).strip("\'") or .strip("'") but the ' ' are still wrapping the filename. How can I remove these?
Happy to provide more details.

It might help to set 'application/pdf' as the mimetype - some mail clients rely on the mimetype to work out which application should open the attachment.
# Additional imports
from email.mime.application import MIMEApplication
import mimetypes
...
for file in filelist:
mimetype, _ = mimetypes.guess_type(file)
mimetype = 'application/octet-stream' if mimetype is None else mimetype
_, _, subtype = mimetype.partition('/')
attachment_part = MIMEApplication(open(file, "rb").read(), subtype)
attachment_part.add_header("Content-Disposition", "attachment; filename='%s'" % os.path.basename(file))
message.attach(attachment_part)
...

Select a random file from a directory and send it (Python, MIME)

I am working on a python program that randomly selects a file from a directory and then sends it to you using the email.mimemodule. I am having a problem where I can choose the random file but I can't sent it due to this error:
File "C:\Users\Mihkel\Desktop\dnak.py", line 37, in sendmemeone
attachment =open(filename, 'rb')
TypeError: expected str, bytes or os.PathLike object, not list
Here is the code:
import smtplib
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
from email import encoders
import os
import random
path ='C:/Users/Mihkel/Desktop/memes'
files = os.listdir(path)
index = random.randrange(0, len(files))
print(files[index])
def send():
email_user = 'yeetbotmemes#gmail.com'
email_send = 'miku.rebane#gmail.com'
subject = 'Test'
msg = MIMEMultipart()
msg['From'] = email_user
msg['To'] = email_send
msg['Subject'] = subject
body = 'Here is your very own dank meme of the day:'
msg.attach(MIMEText (body, 'plain'))
filename=files
attachment =open(filename, 'rb')
part = MIMEBase('application','octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition',"attachment;
filename= "+filename)
msg.attach(part)
text = msg.as_string()
server = smtplib.SMTP('smtp.gmail.com',587)
server.starttls()
server.login(email_user,"MY PASSWORD")
server.sendmail(email_user,email_send,text)
server.quit()
I believe it is just getting the filename as the selected random choice, how could I get it to select the file itself?
EDIT: After making the changes recommended I am now getting this error:
File "C:\Users\Mihkel\Desktop\e8re.py", line 29, in send
part.add_header('Content-Disposition',"attachment; filename= "+filename)
TypeError: can only concatenate str (not "list") to str
Seems like this part is still taking in the list, how would I fix that?

You select a random file and then throw it away (well, you print it, then throw it away):
files = os.listdir(path)
index = random.randrange(0, len(files))
print(files[index])
(which BTW you can do with random.choice(files))
and when calling open you pass it the entire files list:
filename = files
attachment = open(filename, 'rb')
Instead, pass open the file you selected:
attachment = open(random.choice(files), 'rb')
But, this still wouldn't work since listdir only returns the filenames and not the full path, so you will need to get it back, preferably with os.path.join:
attachment = open(os.path.join(path, random.choice(files)), 'rb')

attaching an image (or object) from memory to an email in python

I have an image in memory that I've created (using numpy and PIL), and I'd like to attach it to a created email programatically. I know I could save it to the filesystem, and then reload/attach it, but it seems in-efficient: is there a way to just pipe it to the mime attachment without saving?
The save/reload version:
from PIL import Image
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart
...some img creation steps...
msg = MIMEMultipart()
img_fname = '/tmp/temp_image.jpg'
img.save( img_fname)
with open( img_fname, 'rb') as fp:
img_file = MIMEImage( fp.read() )
img_file.add_header('Content-Disposition', 'attachment', filename=img_fname )
msg.attach( img_file)
...add other attachments and main body of email text...

MIMEImage says that the first argument is just "a string containing the raw image data", so you don't have to open() then .read() it from a file.
If you're making it in PIL and there isn't a way to serialize it directly (there might not be, I can't recall), you can use a io.StringIO (or BytesIO...whichever works with what MIMEImage really wants) file-like buffer to save the file, then read it out as a string. Related question. Modernized adapted excerpt:
import io
from email.mime.image import MIMEImage
# ... make some image
outbuf = io.StringIO()
image.save(outbuf, format="PNG")
my_mime_image = MIMEImage(outbuf.getvalue())
outbuf.close()

Django: CSV in email attachment including Unicode characters results in extra linebreak

I have a reporting feature on my site that send CSV attached file by email. I recently noticed that if one of the string included an accent character my attached CSV has extra line break. Strange thing is I don't see any of these extra linebreak if the string doesn't contain any accent.
Code looks a bit like this:
# -*- coding: utf8 -*-
import unicodecsv
from StringIO import StringIO
from django.core.mail import EmailMultiAlternatives
# Generating the CSV
csvfile = StringIO()
writer = unicodecsv.writer(csvfile, encoding='utf-8')
writer.writerow([u'Test', u'Linebreak è'])
writer.writerow([u'Another', u'line'])
# Email
msg = EmailMultiAlternatives(
'csv report',
'Here is your attached report',
'email#from.com',
'email#to.com'
)
msg.attach('your_report.csv', csvfile.getvalue(), 'text/csv')
msg.send()
Opening the file with VIM shows me something like that:
Test,Linebreak è^M
Another,line
In comparison if the CSV rows include :
writer.writerow([u'Test', u'Linebreak'])
writer.writerow([u'Another', u'line'])
The attached CSV will look like that:
Test,Linebreak
Another,line
The getvalue() seems to output the right EOL formater but something seems to happen once the file is attached. Did someone else noticed similar issue?
(Runing Django 1.6 on python 2.7)
Edit: I have found the root of my problem. Turns out I'm using sendgrid for sending my emails, and for some reason their system is adding extra linebreak on my CSV when this one contains an accent...

As per commenter's request, I'll add a solution that involves Python's stamdard SMTP library instead of SendGrid.
As with OP's code, we use CSV data that is unicode. When it's time to prepare the message, we explictly add the data as UTF-8-encoded text attachment, and construct the message object like so:
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
# Write CSV data ...
msg = MIMEMultipart()
msg['Subject'] = subject
msg['From'] = sender
msg['To'] = recipients
msg.preamble = subject + '\n'
# Message body
body = MIMEText(body, 'plain', 'utf-8')
# CSV data
csv = MIMEText(csvfile.getvalue(), 'csv', 'utf-8')
csv.add_header("Content-Disposition", "attachment",
filename=report_filename)
# Add body and attachment to message
msg.attach(body)
msg.attach(csv)
You can read more about MIMEText in the Python library documentation. I find that passing it unicode strings (as opposed to str/bytes) works as long as there is a properly delcared charset.
Also, I have to note that I'm not sure whether the the newline problem was simply solved by using MIMEText attachment, or because of encoding. It's possible that using MIMEText object as attachment in OP's code may solve the problem. I leave experimentation to you, though.

For those who use Sendgrid as an SMTP provider to send you emails and if you noticed a similar issue, I fixed my problem by not using SMTP but the Web API of Sendgrid (via https://github.com/elbuo8/sendgrid-django).
No more extra lines in my CSV reports now!

How to send a zip file as an attachment in python?

I have looked through many tutorials, as well as other question here on stack overflow, and the documentation and explanation are at minimum, just unexplained code. I would like to send a file that I already have zipped, and send it as an attachment. I have tried copy and pasting the code provided, but its not working, hence I cannot fix the problem.
So what I am asking is if anyone knows who to explain how smtplib as well as email and MIME libraries work together to send a file, more specifically, how to do it with a zip file. Any help would be appreciated.
This is the code that everyone refers to:
import smtplib
import zipfile
import tempfile
from email import encoders
from email.message import Message
from email.mime.base import MIMEBase
from email.mime.multipart import MIMEMultipart
def send_file_zipped(the_file, recipients, sender='you#you.com'):
myzip = zipfile.ZipFile('file.zip', 'w')
# Create the message
themsg = MIMEMultipart()
themsg['Subject'] = 'File %s' % the_file
themsg['To'] = ', '.join(recipients)
themsg['From'] = sender
themsg.preamble = 'I am not using a MIME-aware mail reader.\n'
msg = MIMEBase('application', 'zip')
msg.set_payload(zf.read())
encoders.encode_base64(msg)
msg.add_header('Content-Disposition', 'attachment',
filename=the_file + '.zip')
themsg.attach(msg)
themsg = themsg.as_string()
# send the message
smtp = smtplib.SMTP()
smtp.connect()
smtp.sendmail(sender, recipients, themsg)
smtp.close()
I suspect the issue is this code zips a file as well. I don't want to zip anything as I already have a zipped file I would like to send. In either case, this code is poorly documented as well as the python libraries themselves as they provide no insight on anything past img file and text files.
UPDATE: Error I am getting now. I have also updated what is in my file with the code above
Traceback (most recent call last):
File "/Users/Zeroe/Documents/python_hw/cgi-bin/zip_it.py", line 100, in <module>
send_file_zipped('hw5.zip', 'avaldez#oswego.edu')
File "/Users/Zeroe/Documents/python_hw/cgi-bin/zip_it.py", line 32, in send_file_zipped
msg.set_payload(myzip.read())
TypeError: read() takes at least 2 arguments (1 given)

I don't really see the problem. Just omit the part which creates the zip file and, instead, just load the zip file you have.
Essentially, this part here
msg = MIMEBase('application', 'zip')
msg.set_payload(zf.read())
encoders.encode_base64(msg)
msg.add_header('Content-Disposition', 'attachment',
filename=the_file + '.zip')
themsg.attach(msg)
creates the attachment. The
msg.set_payload(zf.read())
sets, well, the payload of the attachment to what you read from the file zf (probably meaning zip file).
Just open your zip file beforehand and let this line read from it.

I agree the email package is not well documented yet. I investigated it before and wrote a wrapper module that simplifies these kinds of tasks. For example, the following works:
from pycopia import ezmail
# Get the data
data = open("/usr/lib64/python2.7/test/zipdir.zip").read()
# Make a proper mime message object.
zipattachement = ezmail.MIMEApplication.MIMEApplication(data, "zip",
filename="zipdir.zip")
# send it.
ezmail.ezmail(["Here is the zip file.", zipattachement],
To="me#mydomain.com", From="me#mydomain.com", subject="zip send test")
And that's all you need once you have everything installed and configured. :-)

My answer uses shutil to zip a directory containing the attachments and then adds the .zip to the email.
# Importing Dependencies
from email.mime.multipart import MIMEMultipart
from email.mime.application import MIMEApplication
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders
import smtplib
import shutil
def Send_Email():
# Create a multipart message
msg = MIMEMultipart()
Body = MIMEText( {Enter Email Body as str here} )
# Add Headers
msg['Subject'] = ''
msg['From'] = ''
msg['To'] = ''
msg['CC'] = ''
msg['BCC'] = ''
# Add body to email
msg.attach(Body)
# Using Shutil to Zip a Directory
dir_name = {Add Path of the Directory to be Zipped}
output_filename = {Add Output Zip File Path}
shutil.make_archive(output_filename, 'zip', dir_name)
part = MIMEBase("application", "octet-stream")
part.set_payload(open(output_filename + ".zip", "rb").read())
encoders.encode_base64(part)
part.add_header("Content-Disposition", "attachment; filename=\"%s.zip\"" % (output_filename))
msg.attach(part)