CODEWITHSUNDEEP
pythondjango
Caroline
Caroline

Reputation: 739

Django download a file

I'm quite new to using Django and I am trying to develop a website where the user is able to upload a number of excel files, these files are then stored in a media folder Webproject/project/media.

def upload(request):
    if request.POST:
        form = FileForm(request.POST, request.FILES)
        if form.is_valid():
            form.save()
            return render_to_response('project/upload_successful.html')
    else:
        form = FileForm()
    args = {}
    args.update(csrf(request))
    args['form'] = form

    return render_to_response('project/create.html', args)

The document is then displayed in a list along with any other document they have uploaded, which you can click into and it will displays basic info about them and the name of the excelfile they have uploaded. From here I want to be able to download the same excel file again using the link:

 <a  href="/project/download"> Download Document </a>

My urls are

 urlpatterns = [

              url(r'^$', ListView.as_view(queryset=Post.objects.all().order_by("-date")[:25],
                                          template_name="project/project.html")),
              url(r'^(?P<pk>\d+)$', DetailView.as_view(model=Post, template_name="project/post.html")),
              url(r'^upload/$', upload),
              url(r'^download/(?P<path>.*)$', serve, {'document root': settings.MEDIA_ROOT}),

          ] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

but I get the error, serve() got an unexpected keyword argument 'document root'. can anyone explain how to fix this?

OR

Explain how I can get the uploaded files to to be selected and served using

def download(request):
    file_name = #get the filename of desired excel file
    path_to_file = #get the path of desired excel file
    response = HttpResponse(mimetype='application/force-download')
    response['Content-Disposition'] = 'attachment; filename=%s' % smart_str(file_name)
    response['X-Sendfile'] = smart_str(path_to_file)
    return response

Upvotes: 72

Views: 211009

Answers (14)

Mourad Qqch
Mourad Qqch

Reputation: 459

  1. <a href='/your-download-view/' download>Download</a>

  2. In your view:

from django.http import FileResponse

def download(request):
    # pre-processing, authorizations, etc.
    # ...
    return FileResponse(open(path_to_file, 'rb'), as_attachment=True)

Upvotes: 3

cwhisperer
cwhisperer

Reputation: 1926

If the file is a FileField in the model, this is the way I do it:

    try:
        download_file = PrintingFile.objects.get(pk=kwargs.get('pk_file', 0))
        return FileResponse(download_file.file.open(), as_attachment=True)
    except PrintingFile.DoesNotExist:
        raise Http404

More here

Upvotes: 1

Gaurav Nagar
Gaurav Nagar

Reputation: 381

import mimetypes
from django.http import HttpResponse, Http404

mime_type, _ = mimetypes.guess_type(json_file_path)
    
if os.path.exists(json_file_path):
    with open(json_file_path, 'r') as fh:
        response = HttpResponse(fh, content_type=mime_type)
        response['Content-Disposition'] = "attachment; filename=%s" % 'config.json'
        return response
raise Http404

Upvotes: 2

Dhruvin Deshpande
Dhruvin Deshpande

Reputation: 39

Using the below approach makes everything less secure since any user can access any user's file.

<a  href="/project/download" download> Download Document </a>

Using the below approach makes no sense since Django only handles one requests at the time (unless you are using gunicorn or something else), and believe me, the below approach takes a lot of time to complete.

def download(request, path):
    file_path = os.path.join(settings.MEDIA_ROOT, path)
    if os.path.exists(file_path):
        with open(file_path, 'rb') as fh:
            response = HttpResponse(fh.read(), content_type="application/vnd.ms-excel")
            response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)
            return response
    raise Http404

So what is the optimum solution? Use Nginx authenticated routes. When requesting a file from Nginx you can make a request to a route and depending on the HTTP response Nginx allows to denies that request. This makes it very secure and also scalable and performant. You can ready about more here

Upvotes: 2

MOhammad.Isaac.Ahmadi
MOhammad.Isaac.Ahmadi

Reputation: 379

1.settings.py:

MEDIA_DIR = os.path.join(BASE_DIR,'media')
#Media
MEDIA_ROOT = MEDIA_DIR
MEDIA_URL = '/media/'

2.urls.py:

from django.conf.urls.static import static
urlpatterns += static(settings.MEDIA_URL,document_root=settings.MEDIA_ROOT)

3.in template:

<a href="{{ file.url }}" download>Download File.</a>

Work and test in django >=3

for more detail use this link: https://youtu.be/MpDZ34mEJ5Y

Upvotes: 1

El Bachir
El Bachir

Reputation: 121

@Biswadp's solution worked greatly for me

In your static folder, make sure to have the desired files you would like the user to download

In your HTML template, your code should look like this :

<a href="{% static 'Highlight.docx' %}"> Download </a>

Upvotes: 2

Timeless
Timeless

Reputation: 104

If you hafe upload your file in media than:

media
example-input-file.txt

views.py

def download_csv(request):    
    file_path = os.path.join(settings.MEDIA_ROOT, 'example-input-file.txt')    
    if os.path.exists(file_path):    
        with open(file_path, 'rb') as fh:    
            response = HttpResponse(fh.read(), content_type="application/vnd.ms-excel")    
            response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)    
            return response

urls.py

path('download_csv/', views.download_csv, name='download_csv'),

download.html

a href="{% url 'download_csv' %}" download=""

Upvotes: 0

Shahriar.M
Shahriar.M

Reputation: 898

I use this method:

{% if quote.myfile %}
    <div class="">
        <a role="button" 
            href="{{ quote.myfile.url }}"
            download="{{ quote.myfile.url }}"
            class="btn btn-light text-dark ml-0">
            Download attachment
        </a>
    </div>
{% endif %}

Upvotes: 0

sikaili99
sikaili99

Reputation: 582

When you upload a file using FileField, the file will have a URL that you can use to point to the file and use HTML download attribute to download that file you can simply do this.

models.py

The model.py looks like this

class CsvFile(models.Model):
    csv_file = models.FileField(upload_to='documents')

views.py

#csv upload

class CsvUploadView(generic.CreateView):

   model = CsvFile
   fields = ['csv_file']
   template_name = 'upload.html'

#csv download

class CsvDownloadView(generic.ListView):

    model = CsvFile
    fields = ['csv_file']
    template_name = 'download.html'

Then in your templates.

#Upload template

upload.html

<div class="container">
<form action="#" method="POST" enctype="multipart/form-data">
    {% csrf_token %}
    {{ form.media }}
    {{ form.as_p }}
    <button class="btn btn-primary btn-sm" type="submit">Upload</button>
</form>

#download template

download.html

  {% for document in object_list %}

     <a href="{{ document.csv_file.url }}" download  class="btn btn-dark float-right">Download</a>

  {% endfor %}

I did not use forms, just rendered model but either way, FileField is there and it will work the same.

Upvotes: 9

mananbh9
mananbh9

Reputation: 581

Reference:

In view.py Implement function like,

def download(request, id):
    obj = your_model_name.objects.get(id=id)
    filename = obj.model_attribute_name.path
    response = FileResponse(open(filename, 'rb'))
    return response

Upvotes: 45

Biswadp
Biswadp

Reputation: 21

Simple using html like this downloads the file mentioned using static keyword

<a href="{% static 'bt.docx' %}" class="btn btn-secondary px-4 py-2 btn-sm">Download CV</a>

Upvotes: 1

Hasan Basri
Hasan Basri

Reputation: 868

You can add "download" attribute inside your tag to download files.

<a  href="/project/download" download> Download Document </a>

https://www.w3schools.com/tags/att_a_download.asp

Upvotes: 66

Sergey Gornostaev
Sergey Gornostaev

Reputation: 7807

You missed underscore in argument document_root. But it's bad idea to use serve in production. Use something like this instead:

import os
from django.conf import settings
from django.http import HttpResponse, Http404

def download(request, path):
    file_path = os.path.join(settings.MEDIA_ROOT, path)
    if os.path.exists(file_path):
        with open(file_path, 'rb') as fh:
            response = HttpResponse(fh.read(), content_type="application/vnd.ms-excel")
            response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)
            return response
    raise Http404

Upvotes: 139

Don Kirkby
Don Kirkby

Reputation: 56230

I've found Django's FileField to be really helpful for letting users upload and download files. The Django documentation has a section on managing files. You can store some information about the file in a table, along with a FileField that points to the file itself. Then you can list the available files by searching the table.

Upvotes: 4

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