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)
Related
So first of all I'm not really sure if I'm going about attaching this zip folder the right way. I've kinda hacked together different things together, but I'm getting an error message below which I've linked towards the bottom of my code.
from email.message import EmailMessage
import shutil
import smtplib
import base64
message = EmailMessage()
message["From"] = "superdummy#idiot.com"
message["To"] = "superdummy#idiot.com"
message["Subject"] = "Testing Zip"
path = "Testing.zip"
with open(path, "rb") as f:
bytes = f.read()
encoded = base64.b64encode(bytes)
message.add_attachment(encoded, maintype='application/zip', subtype='octet-stream', filename="Testing.zip")
s = smtplib.SMTP(host='blah.blah.yum', port=99)
s.send_message(message)
When I receive the email and I try to open it up, I get the error message:
"Windows cannot open the folder, the compressed Zip folder is invalid"
Couldn't find similar error messages on Stackoverflow or other places. I'd really appreciate some direction. Thanks!
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)
...
I have a script that runs main() and at the end I want to send the contents it has by e-mail. I don't want to write new files nor anything. Just have the original script be unmodified and at the end just send the contents of what it printed. Ideal code:
main()
send_mail()
I tried this:
def main():
print('HELLOWORLD')
def send_email(subject='subject', message='', destination='me#gmail.com', password_path=None):
from socket import gethostname
from email.message import EmailMessage
import smtplib
import json
import sys
server = smtplib.SMTP('smtp.gmail.com', 587)
smtplib.stdout = sys.stdout # <<<<<-------- why doesn't it work?
server.starttls()
with open(password_path) as f:
config = json.load(f)
server.login('me#gmail.com', config['password'])
# craft message
msg = EmailMessage()
#msg.set_content(message)
msg['Subject'] = subject
msg['From'] = 'me#gmail.com'
msg['To'] = destination
# send msg
server.send_message(msg)
if __name__ == '__main__':
main()
send_mail()
but it doesn't work.
I don't want to write other files or change the original python print statements. How to do this?
I tried this:
def get_stdout():
import sys
print('a')
print('b')
print('c')
repr(sys.stdout)
contents = ""
#with open('some_file.txt') as f:
#with open(sys.stdout) as f:
for line in sys.stdout.readlines():
contents += line
print(contents)
but it does not let me read sys.stdout because it says its not readable. How can I open it in readable or change it to readable in the first place?
I checked all of the following links but none helped:
How to send output from a python script to an email address
https://www.quora.com/How-can-I-send-an-output-from-a-Python-script-to-an-email-address
https://bytes.com/topic/python/answers/165835-email-module-redirecting-stdout
Redirect stdout to a file in Python?
How to handle both `with open(...)` and `sys.stdout` nicely?
Capture stdout from a script?
To send e-mails I am using:
def send_email(subject, message, destination, password_path=None):
from socket import gethostname
from email.message import EmailMessage
import smtplib
import json
server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
with open(password_path) as f:
config = json.load(f)
server.login('me123#gmail.com', config['password'])
# craft message
msg = EmailMessage()
message = f'{message}\nSend from Hostname: {gethostname()}'
msg.set_content(message)
msg['Subject'] = subject
msg['From'] = 'me123#gmail.com'
msg['To'] = destination
# send msg
server.send_message(msg)
note I have my password in a json file using an app password as suggested by this answer https://stackoverflow.com/a/60996409/3167448.
using this to collect the contents from stdout by writing it to a custom stdout file using the builtin function print:
import sys
from pathlib import Path
def my_print(*args, filepath='~/my_stdout.txt'):
filepath = Path(filepath).expanduser()
# do normal print
__builtins__['print'](*args, file=sys.__stdout__) #prints to terminal
# open my stdout file in update mode
with open(filepath, "a+") as f:
# save the content we are trying to print
__builtins__['print'](*args, file=f) #saves in a file
def collect_content_from_file(filepath):
filepath = Path(filepath).expanduser()
contents = ''
with open(filepath,'r') as f:
for line in f.readlines():
contents = contents + line
return contents
Note the a+ to be able to create the file if it already does NOT exist.
Note that if you want to delete the old contents of your custom my_stdout.txt you need to delete the file and check if it exists:
# remove my stdout if it exists
os.remove(Path('~/my_stdout.txt').expanduser()) if os.path.isfile(Path('~/my_stdout.txt').expanduser()) else None
The credits for the print code are from the answer here: How does one make an already opened file readable (e.g. sys.stdout)?
I am newbie for python and had a task to send multiple email with relevant attachments. I will eloborate it , A folder contains multiple pdf files each file contains some text including email id.I need to read the email id from the each pdf file and send the same file as the attachment to the mailid in the pdf file. below is the code for reference
# Get the count of files in the folder
import os
import re
global str
import smtplib
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
cpt = sum([len(files) for r, d, files in
os.walk("D:\MyOfficeDocuments\ADCB\PythonScripts\PdfFiles")])
#Reading Mail from each pdf file and send the same file as attachment to
these mails
import PyPDF2
from os import listdir
from os.path import isfile, join
from PyPDF2 import PdfFileWriter, PdfFileReader
mypath='D:\MyOfficeDocuments\ADCB\PythonScripts\PdfFiles'
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]
for file in onlyfiles:
count = 1
while count <cpt:
os.chdir(r'D:\MyOfficeDocuments\ADCB\PythonScripts\PdfFiles')
pdfFileObj = open(file,'rb')
pdfReader = PyPDF2.PdfFileReader(pdfFileObj)
pageObj = pdfReader.getPage(0)
count +=1
text = pageObj.extractText()
email_user='madhugut82#gmail.com'
eline = re.findall('\S+#\S+.com', text)
email_send=eline
print(file)
password='harshi54537'
subject='Python !'
msg=MIMEMultipart()
msg['From']=email_user
msg['To']=', '.join(email_send)
#listalink = " ".join(listalink)
msg['Subject']=subject
#print (email_send)
body='Hi there, sending this email from python using python scripting'
msg.attach(MIMEText(body,'plain'))
filename
='D:\MyOfficeDocuments\ADCB\PythonScripts\Destination\Document.txt'
attachment=open(file,'rb')
#print(attachment)
part=MIMEBase('application','pdf')
part.set_payload(attachment.read())
part.add_header('Content-Disposition',"attachement; filename="+file)
msg.attach(part)
#email.encoders.encode_base64(part)
print('x')
text=msg.as_string()
#text=msg.encode("utf8")
#text=msg.as_string().encode('utf-8','ignore')
#text=msg.as_string().encode('ascii','ignore')
server=smtplib.SMTP('smtp.gmail.com',587)
server.starttls()
server.login(email_user,password)
server.sendmail(email_user,email_send,text)
#server.sendmail(email_user,email_send,msg.encode("utf8"))
server.quit()
As per the above code I am getting an error message shown in below
msg = _fix_eols(msg).encode('ascii')
UnicodeEncodeError: 'ascii' codec can't encode characters in position 559-562: ordinal not in range(128)
But if I change the code as below for
text=ms.as_string().encode("UTF")
I am not getting any error but attachement is showing blank
Please suggest me where the exact issue is and what is the issue to getting blank pdf attachment.
I requesting you if there is any code suggestion then please suggest for pdf file only
Thanks in Advance
Madhu
Your problem is that you are using a simple MIMEBase for the (binary) pdf file. As MIMEBase is a parent class for various possible message type, it does not encode its payload, and your message contains raw 8 bits bytes.
Two possible fixes here:
just base64 encode the pdf file content:
...
from email.encoders import encode_base64
...
part=MIMEBase('application','pdf')
part.set_payload(attachment.read())
part.add_header('Content-Disposition',"attachement; filename="+file)
encode_base64(part)
msg.attach(part)
...
use the more specialized MIMEApplication which encodes everything by default:
...
from email.mime.application import MIMEApplication
...
part=MIMEApplication(attachment.read(),'pdf')
part.add_header('Content-Disposition',"attachement; filename="+file)
msg.attach(part)
...
I advise you to use that second way, because the documentation for MIMEBase says:
Ordinarily you won’t create instances specifically of MIMEBase, although you could. MIMEBase is provided primarily as a convenient base class for more specific MIME-aware subclasses.
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')