Develop Reference

how can I download files from the URL that does not have file name in it and how to get to know its file extension or file type

this is an URL example "https://procurement-notices.undp.org/view_file.cfm?doc_id=257280"
if you put it in the browser a file will start downloading in your system.
I want to download this file using python and store it somewhere on my computer
this is how tried
import requests
# first_url = 'https://readthedocs.org/projects/python-guide/downloads/pdf/latest/'
second_url="https://procurement-notices.undp.org/view_file.cfm?doc_id=257280"
myfile = requests.get(second_url , allow_redirects=True)
# this works for the first URL
# open('example.pdf' , 'wb').write(myfile.content)
# this did't work for both of them
# open('example.txt' , 'wb').write(myfile.content)
# this works for the second URL
open('example.doc' , 'wb').write(myfile.content)
first: if I put the first_url in the browser it will download a pdf file, putting second_url will download a .doc file How can I know what type of file will the URL give to us or what type of file will be downloaded so that I use the correct open(...) method?
second: If I use the second URL in the browser a file with the name "T__proc_notices_notices_080_k_notice_doc_79545_770020123.docx" starts downloading. how can I know this file name when I try to download the file?
if you know any better solution kindly let me know for the implementation.
kindly have a quick look at Downloading Files from URLs and zip downloaded files in python question aswell

myfile.headers['content-type'] will give you the MIME-type of the URL's content and myfile.headers['content-disposition'] gives you info like filename etc. (if the response contains this header at all)

you can use response headers content-type like for first url it is application/pdf and sencond url for is application/msword you save file according to it. you can make extension dictinary where you can store possible file format and their types and match with it. your second question is also same like this one so i am taking your two urls from that question and for file name i am using just integers
all_Urls = ['https://omextemplates.content.office.net/support/templates/en-us/tf16402488.dotx' ,
'https://procurement-notices.undp.org/view_file.cfm?doc_id=257280']
extension_dict = {'application/vnd.openxmlformats-officedocument.wordprocessingml.document':'.docx',
'application/vnd.openxmlformats-officedocument.wordprocessingml.template':'.dotx',
'application/vnd.ms-word.document.macroEnabled.12':'.docm',
'application/vnd.ms-word.template.macroEnabled.12':'.dotm',
'application/pdf':'.pdf',
'application/msword':'.doc'}
for i,url in enumerate(all_Urls):
resp = requests.get(url)
response_headers = resp.headers
file_extension = extensio_dict[response_headers['Content-Type']]
with open(f"{i}.{file_extension}",'wb') as f:
f.write(resp.content)
for MIME-Type see this answer

How to send csv file as an attachment using python while taking care of leading zeros

I want to send a email using CSV File as an attachment ,When I open the email I receive an attachment where we do not have a Leading Zero ,I want leading zero to appear in the attachment of CSV File using python ,Please suggest me a workaround .
Results file is the CSV File :
The sample data looks : 0763808075
But when I run the python script the email attachment the data looks like this : 763808075.
I want the leading zero to be added in the email attachment
with open ( resultsfile1 , "rb" ) as attached_file :
part = MIMEApplication (
attached_file.read() ,
Name=resultsfile1
)
# After the file is closed
result= os.path.basename ('\filename.csv')
part['Content-Disposition'] = 'attachment; filename={} '.format(str(result))
msg.attach ( part )

Download zip file with Django

I'm quite new on Django and i'm looking for a way to dwonload a zip file from my django site but i have some issue when i'm running this piece of code:
def download(self):
dirName = settings.DEBUG_FOLDER
name = 'test.zip'
with ZipFile(name, 'w') as zipObj:
# Iterate over all the files in directory
for folderName, subfolders, filenames in os.walk(dirName):
for filename in filenames:
# create complete filepath of file in directory
filePath = os.path.join(folderName, filename)
# Add file to zip
zipObj.write(filePath, basename(filePath))
path_to_file = 'http://' + sys.argv[-1] + '/' + name
resp= {}
# Grab ZIP file from in-memory, make response with correct MIME-type
resp = HttpResponse(content_type='application/zip')
# ..and correct content-disposition
resp['Content-Disposition'] = 'attachment; filename=%s' % smart_str(name)
resp['X-Sendfile'] = smart_str(path_to_file)
return resp
I get:
Exception Value:
<HttpResponse status_code=200, "application/zip"> is not JSON serializable
I tried to change the content_type to octet-stream but it doesn't work
And to use a wrapper as followw:
wrapper = FileWrapper(open('test.zip', 'rb'))
content_type = 'application/zip'
content_disposition = 'attachment; filename=name'
# Grab ZIP file from in-memory, make response with correct MIME-type
resp = HttpResponse(wrapper, content_type=content_type)
# ..and correct content-disposition
resp['Content-Disposition'] = content_disposition
I didn't find useful answer so far but maybe I didn't search well, so if it seems my problem had been already traited, feel free to notify me
Thank you very much for any help

You have to send the zip file as byte
response = HttpResponse(zipObj.read(), content_type="application/zip")
response['Content-Disposition'] = 'attachment; filename=%s' % smart_str(name)
return response

I would do like this:
(Caveat I use wsl so the python function will make use of cmd lines)
In view:
import os
def zipdownfun(request):
""" Please establish in settings.py where media file should be downloaded from.
In my case is media with a series of other folders inside. Media folder is at the same level of project root folder, where settings.py is"""
file_name = os.path.join(MEDIA_URL,'folder_where_your_file_is','file_name.zip')
"""let us put the case that you have zip folder in media folder"""
file_folder_path = os.path.join(MEDIA_URL,'saving_folder')
"""The command line takes as first variable the name of the
future zip file and as second variable the destination folder"""
cmd = f'zip {file_name} {file_folder_path}'
"""With os I open a process in the background so that some magic
happens"""
os.system(cmd)
"""I don't know what you want to do with this, but I placed the
URL of the file in a button for the download, so you will need
the string of the URL to place in href of an <a> element"""
return render(request,'your_html_file.html', {'url':file_name})
The db I have created, will be updated very often. I used a slightly different version of this function with -r clause since I had to zip, each time, a folder. Why I did this? The database I have created has to allow the download of this zipped folder. This folder will be updated daily. So this function basically overwrites the file each time that is downloaded. It will be so fresh of new data each time.
Please refer to this page to understand how to create a button for the download of the generated file.
Take as reference approach 2. The URL variable that you are passing to the Django template should be used at the place of the file (screenshot attached)
I hope it can help!

Parse excel attachment from .eml file in python

I'm trying to parse a .eml file. The .eml has an excel attachment that's currently base 64 encoded. I'm trying to figure out how to decode it into XML so that I can later turn it into a CSV I can do stuff with.
This is my code right now:
import email
data = file('Openworkorders.eml').read()
msg = email.message_from_string(data)
for part in msg.walk():
c_type = part.get_content_type()
c_disp = part.get('Content Disposition')
if part.get_content_type() == 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet':
excelContents = part.get_payload(decode = True)
print excelContents
The problem is
When I try to decode it, it spits back something looking like this.
I've used this post to help me write the code above.
How can I get an email message's text content using Python?
Update:
This is exactly following the post's solution with my file, but part.get_payload() returns everything still encoded. I haven't figured out how to access the decoded content this way.
import email
data = file('Openworkorders.eml').read()
msg = email.message_from_string(data)
for part in msg.walk():
if part.get_content_type() == 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet':
name = part.get_param('name') or 'MyDoc.doc'
f = open(name, 'wb')
f.write(part.get_payload(None, True))
f.close()
print part.get("content-transfer-encoding")

As is clear from this table (and as you have already concluded), this file is an .xlsx. You can't just decode it with unicode or base64: you need a special package. Excel files specifically are a bit tricker (for e.g. this one does PowerPoint and Word, but not Excel). There are a few online, see here - xlrd might be the best.

Here is my solution:
I found 2 things out:
1.) I thought .open() was going inside the .eml and changing the selected decoded elements. I thought I needed to see decoded data before moving forward. What's really happening with .open() is it's creating a new file in the same directory of that .xlsx file. You must open the attachment before you will be able to deal with the data.
2.) You must open an xlrd workbook with the file path.
import email
import xlrd
data = file('EmailFileName.eml').read()
msg = email.message_from_string(data) # entire message
if msg.is_multipart():
for payload in msg.get_payload():
bdy = payload.get_payload()
else:
bdy = msg.get_payload()
attachment = msg.get_payload()[1]
# open and save excel file to disk
f = open('excelFile.xlsx', 'wb')
f.write(attachment.get_payload(decode=True))
f.close()
xls = xlrd.open_workbook(excelFilePath) # so something in quotes like '/Users/mymac/thisProjectsFolder/excelFileName.xlsx'
# Here's a bonus for how to start accessing excel cells and rows
for sheets in xls.sheets():
list = []
for rows in range(sheets.nrows):
for col in range(sheets.ncols):
list.append(str(sheets.cell(rows, col).value))

Save an django email to eml

I am generating a bunch of html emails in django, and I want to save them into a model, in a FileField. I can quite easily generate the html content and dump in into a File, but I want to create something that can be opened in email clients, e.g. an eml file. Does anyone know of a python or django module to do this? Just to be clear, I'm not looking for an alternative email backend, as I also want the emails to be sent when they're generated.
Edit: After a bit of reading, it looks to me like the EmailMessage.messge() should return the content that should be stored int he eml file. However, if I try to save it like this, the file generated is empty:
import tempfile
name = tempfile.mkstemp()[1]
fh = open(name, 'wb')
fh.write(bytes(msg.message()))
fh.close()
output = File(open(name, 'rb'), msg.subject[:50])
I want to use a BytesIO instead of a temp file, but the temp file is easier for testing.

EML file is actually a text file with name value pairs. A valid EML file would be like
From: test#example.com
To: test#example.com
Subject: Test
Hello world!
If you follow the above pattern and save it in file with .eml extension, thunderbird like email clients will parse and show them without any problem.

Django's EmailMessage.message().as_bytes() will return the content of the .eml file. Then you just need to save the file to the directory of your choice:
from django.core.mail import EmailMessage
msg = EmailMessage(
'Hello',
'Body goes here',
'from#example.com',
['to3#example.com'],
)
eml_content = msg.message().as_bytes()
file_name = "/path/to/eml_output.eml"
with open(file_name, "wb") as outfile:
outfile.write(eml_content)

I had the similar problem. I found ticket on Django site. Last comment suggests using django-eml-email-backend. It helps me and it is very useful and simple.
Example:
installing:
$ pip install django-eml-email-backend
using:
EMAIL_BACKEND = 'eml_email_backend.EmailBackend'
EMAIL_FILE_PATH = 'path/to/output/folder/'