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django error:can't specify the download filename

def download(request):
f = open("next_op.xls")
data = f.read()
f.close()
response = HttpResponse(data, content_type = './application/vnd.ms-excel')
response['Content-Disposition'] = 'attachment; filename="nextop.xls"'
return response
When I use this code, I can download the file correctly, but the file name is invalid. I get the file name "download", and I found the response header doesn't include the Content-Disposition after I download the file.

you may need this:
fd = open(file, 'rb')
return FileResponse(fd, as_attachment=True, filename='xxx')

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.

Unknown and Damaged Archive in django

This is code to download files but when file downloads and i open them :the archive is unknown and damaged. Can you please help me to solve this problem here code is:
def download(request):
file_name =request.GET.get('file_name', '')
the_file = "C:\\Users\\CV_Uploads\\uploadfiles\\uploadfiles\\uploaded_files\\1395901478_89_uploadfiles.rar"
filename = os.path.basename(the_file)
response = HttpResponse(FileWrapper(open(the_file)),
content_type=mimetypes.guess_type(the_file)[0])
response['Content-Length'] = os.path.getsize(the_file)
response['Content-Disposition'] = "attachment; filename=%s" % filename
return response

When you are dealing with paths you should use raw string .
use
the_file = r"C:\Users\CV_Uploads\uploadfiles\uploadfiles\uploaded_files\1395901478_89_uploadfiles.rar"

Download files from server in web2py

I am currently setting up a website where I get a file uploaded from the user , do some processing on it and provide a link for the user to download the processed file from. I presently want to provide a path to the file on my local system, I am new to web2py, and am having trouble doing this.
Could someone please help me do this?
Regards

see this link for some hint: webpy: how to stream files , and may be add some code like this:
BUF_SIZE = 262144
class download:
def GET(self):
file_name = # get from url
file_path = os.path.join('/path to your file', file_name)
f = None
try:
f = open(file_path, "rb")
webpy.header('Content-Type','application/octet-stream')
webpy.header('Content-disposition', 'attachment; filename=%s' % file_name)
while True:
c = f.read(BUF_SIZE)
if c:
yield c
else:
break
except Exception, e:
# throw 403 or 500 or just leave it
pass
finally:
if f:
f.close()