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Django: How do I download .xls file through a django view

I have a button which downloads a excel file with extension .xls. I am using module xlrd to parse the file and return it back to the user. However it appears to add the object name into the excel file instead of the data.
How can I return the file to the user with the data rather than the objects name?
View
def download_file(self, testname):
import csv, socket, os, xlrd
extension = '.xls'
path = r"C:\tests\{}_Report{}".format(testname, extension)
try:
f = xlrd.open_workbook(path)
response = HttpResponse(f, content_type='application/ms-excel')
response['Content-Disposition'] = 'attachment; filename={}_Report{}'.format(testname, extension)
return response
except Exception as Error:
return HttpResponse(Error)
return redirect('emissions_dashboard:overview_view_record')
Excel result
Download successful:
Content:
Note: I understand this is an old file format but is required for this particular project.

You are trying to send a xlrd.book.Book object, not a file.
You used xlrd to do your things in the workbook, and then saved to a file.
workbook = xlrd.open_workbook(path)
#... do something
workbook.save(path)
Now you send it like any other file:
with open(path, 'rb') as f:
response = HttpResponse(f.read(), content_type="application/ms-excel")
response['Content-Disposition'] = 'attachment; filename={}_Report{}'.format(testname, extension)

Response ZIP file by Django

I try to download img from url, add it to zip archive and then response this archive by Django HttpResponse.
import os
import requests
import zipfile
from django.http import HttpResponse
url = 'http://some.link/img.jpg'
file = requests.get(url)
data = file.content
rf = open('tmp/pic1.jpg', 'wb')
rf.write(data)
rf.close()
zipf = zipfile.ZipFile('tmp/album.zip', 'w') # This file is ok
filename = os.path.basename(os.path.normpath('tmp/pic1.jpg'))
zipf.write('tmp/pic1.jpg', filename)
zipf.close()
resp = HttpResponse(open('tmp/album.zip', 'rb'))
resp['Content-Disposition'] = 'attachment; filename=album.zip'
resp['Content-Type'] = 'application/zip'
return resp # Got corrupted zip file
When I save file to tmp folder - it's ok, I can extract it.
But when I response this file I get 'Error 1/2/21' on MacOS or Unexpected EOF if I try to open in Atom editor (just for test).
I also used StringIO instead of saving zip file, but it doesn't influence the result.

If you're using Python 3, you'd do it like this:
import os, io, zipfile, requests
from django.http import HttpResponse
# Get file
url = 'https://some.link/img.jpg'
response = requests.get(url)
# Get filename from url
filename = os.path.split(url)[1]
# Create zip
buffer = io.BytesIO()
zip_file = zipfile.ZipFile(buffer, 'w')
zip_file.writestr(filename, response.content)
zip_file.close()
# Return zip
response = HttpResponse(buffer.getvalue())
response['Content-Type'] = 'application/x-zip-compressed'
response['Content-Disposition'] = 'attachment; filename=album.zip'
return response
That's without saving the file. Downloaded file goes directly to io.
To response saved file, use this syntax:
response = HttpResponse(open('path/to/file', 'rb').read())

Downloading file from FileField in Django with a HTTP link in a HTML file

I create a link that when a user press it, it will download a pdf file from the media folder in Django to the users machine.
I tried different methods but all were wrong for me. It tells me that the file can not be found, or the code is running but the file is corrupted.
My Html link:
<td> Download</td>
My url pattern links into a view:
url(r'^download/$', views.DownloadPdf),
My FileField is like this:
upload_pdf = models.FileField()
Following snippet code is the view that downloads a corrupted pdf:
def DownloadPdf(request):
filename = '/home/USER/PycharmProjects/MyProject/media/Invoice_Template.pdf'
response = HttpResponse(content_type='application/pdf')
fileformat = "pdf"
response['Content-Disposition'] = 'attachment;
filename=thisismypdf'.format(fileformat)
return response
So, what I have to do to make it working ?

with open(os.path.join(settings.MEDIA_ROOT, 'Invoice_Template.pdf'), 'rb') as fh:
response = HttpResponse(fh.read(), content_type="application/pdf")
response['Content-Disposition'] = 'attachment; filename=invoice.pdf'
return response

#Sergey Gornostaev 's code just work perfect, but i post below my code because it is a aproach in a differect way.
I correct a little bit my code:
def DownloadPdf(request):
path_to_file = '/home/USER/PycharmProjects/MyProject/media /Invoice_Template.pdf'
f = open(path_to_file, 'r')
myfile = File(f)
response = HttpResponse(myfile, content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename=filename'
return response
But only in pdf file ( works with txt files ) gives me that error:
'utf-8' codec can't decode byte 0xe2 in position 10

Python - create txt file on the fly and send it via FTP?

So i am currently creating a text file from a jinja2 template on the fly and having it be downloaded by the users browser, however i want to add an option to send it somewhere via FTP (all the FTP details are predefined and wont change)
how do i create the file to be sent?
Thanks
code:
...
device_config.stream(
STR = hostname,
IP = subnet,
BGPASNO = bgp_as,
LOIP = lo1,
DSLUSER = dsl_user,
DSLPASS = dsl_pass,
Date = install_date,
).dump(config_file)
content = config_file.getvalue()
content_type = 'text/plain'
content_disposition = 'attachment; filename=%s' % (file_name)
response = None
if type == 'FILE':
response = HttpResponse(content, content_type=content_type)
response['Content-Disposition'] = content_disposition
elif type == 'FTP':
with tempfile.NamedTemporaryFile() as temp:
temp.write(content)
temp.seek(0)
filename = temp.name
session = ftplib.FTP('192.168.1.1','test','password')
session.storbinary('STOR {0}'.format(file_name), temp)
session.quit()
temp.flush()
return response
EDIT
needed to add temp.seek(0) before sending the file

You can use the tempfile module to create a named temporary file.
import tempfile
with tempfile.NamedTemporaryFile() as temp:
temp.write(content)
temp.flush()
filename = temp.name
session.storbinary('STOR {0}'.format(file_name), temp)

Here is a working example using BytesIO under io module. Code is tested and works.
import ftplib
import io
session = ftplib.FTP('192.168.1.1','USERNAME','PASSWORD')
# session.set_debuglevel(2)
buf=io.BytesIO()
# b'str' to content of buff.write() as it throws an error in python3.7
buf.write(b"test string")
buf.seek(0)
session.storbinary("STOR testfile.txt",buf)
session.quit()

How to open a .txt file in Django?

I have a Django app in which when I click on the link then I can download a .txt file. Now instead of downloading that file I need to open that file (in 'r' mode). I'm trying to do something similar to that of mail attachments that is when we click on the attachment then it opens up instead of downloading. How can I do it ? The following code is to download the .txt file :
def fetch_logfile(request,logfile):
try:
folder,log,_ = logfile.split("/")
pathRelative = r"/LogFile/"+log
folder,log,_ = logfile.split("/")
pathRelative = r"/LogFile/"+log
path = pathRelative[1::]
os.startfile(pathRelative,open)
file_path =os.getcwd()+ '/' +pathRelative
file_wrapper = FileWrapper(file(file_path,'rb'))
file_mimetype = mimetypes.guess_type(file_path)
response = HttpResponse(file_wrapper, content_type=file_mimetype )
response['X-Sendfile'] = file_path
response['Content-Length'] = os.stat(file_path).st_size
nameOnly = log.split('/')
response['Content-Disposition'] = 'attachment; filename=%s' % nameOnly[len(nameOnly)-1]
return response
except:
## do something else
The following code works which I have tried in Python IDLE but when I try the same in Django then it doesn't work. I'm not sure if this is the right way either.Please advice me on this.
def fetch_Logfile(request,logfile):
import os,sys
path = "C:\\Users\\welcome\\Desktop\\mysite\\LogFile\\"+"756849.txt"
os.startfile(path,open)
## do something with logfile and request
def fetch_Logfile(request,logfile):
path = "C:\\Users\\welcome\\Desktop\\mysite\\LogFile\\"+"756849.txt"
import webbrowser
webbrowser.open(path)
## do something with logfile and request
def fetch_Logfile(request,logfile):
import win32api,os,subprocess
path = "C:\\Users\\welcome\\Desktop\\mysite\\LogFile\\"+"756849.txt"
filename_short = win32api.GetShortPathName(path)
subprocess.Popen('start ' + filename_short, shell=True )
subprocess.Popen('start ' + path, shell=True )
## do something with logfile and request

my_file = open(file_path, 'r')
response = HttpResponse(my_file.read(), mimetype='text/plain')
response['Content-Disposition'] = 'inline;filename=some_file.txt'
return response
Here is the MIME Types – Complete List
You can provide mimetype = ' / ' based on the file extension by referencing the mime type list.