I created a custom group in Django’s admin site.

In my code, I want to check if a user is in this group. How do I do that?

You can access the groups simply through the groups attribute on User.

from django.contrib.auth.models import User, Group

group = Group(name = "Editor")
group.save()                    # save this new group for this example
user = User.objects.get(pk = 1) # assuming, there is one initial user 
user.groups.add(group)          # user is now in the "Editor" group

then user.groups.all() returns [<Group: Editor>].

Alternatively, and more directly, you can check if a a user is in a group by:

if django_user.groups.filter(name = groupname).exists():

    ...

Note that groupname can also be the actual Django Group object.

Your User object is linked to the Group object through a ManyToMany relationship.

You can thereby apply the filter method to user.groups.

So, to check if a given User is in a certain group (“Member” for the example), just do this :

def is_member(user):
    return user.groups.filter(name='Member').exists()

If you want to check if a given user belongs to more than one given groups, use the __in operator like so :

def is_in_multiple_groups(user):
    return user.groups.filter(name__in=['group1', 'group2']).exists()

Note that those functions can be used with the @user_passes_test decorator to manage access to your views :

from django.contrib.auth.decorators import login_required, user_passes_test
@login_required
@user_passes_test(is_member) # or @user_passes_test(is_in_multiple_groups)
def myview(request):
    # Do your processing

Hope this help

If you need the list of users that are in a group, you can do this instead:

from django.contrib.auth.models import Group
users_in_group = Group.objects.get(name="group name").user_set.all()

and then check

 if user in users_in_group:
     # do something

to check if the user is in the group.

If you don’t need the user instance on site (as I did), you can do it with

User.objects.filter(pk=userId, groups__name='Editor').exists()

This will produce only one request to the database and return a boolean.

If a user belongs to a certain group or not, can be checked in django templates using:

{% if group in request.user.groups.all %} "some action" {% endif %}

You just need one line:

from django.contrib.auth.decorators import user_passes_test  

@user_passes_test(lambda u: u.groups.filter(name='companyGroup').exists())
def you_view():
    return HttpResponse("Since you're logged in, you can see this text!")

Just in case if you wanna check user’s group belongs to a predefined group list:

def is_allowed(user):
    allowed_group = set(['admin', 'lead', 'manager'])
    usr = User.objects.get(username=user)
    groups = [ x.name for x in usr.groups.all()]
    if allowed_group.intersection(set(groups)):
       return True
    return False

I have similar situation, I wanted to test if the user is in a certain group. So, I’ve created new file utils.py where I put all my small utilities that help me through entire application. There, I’ve have this definition:

utils.py

def is_company_admin(user):
    return user.groups.filter(name='company_admin').exists()

so basically I am testing if the user is in the group company_admin and for clarity I’ve called this function is_company_admin.

When I want to check if the user is in the company_admin I just do this:

views.py

from .utils import *

if is_company_admin(request.user):
        data = Company.objects.all().filter(id=request.user.company.id)

Now, if you wish to test same in your template, you can add is_user_admin in your context, something like this:

views.py

return render(request, 'admin/users.html', {'data': data, 'is_company_admin': is_company_admin(request.user)})

Now you can evaluate you response in a template:

users.html

{% if is_company_admin %}
     ... do something ...
{% endif %}

Simple and clean solution, based on answers that can be found earlier in this thread, but done differently. Hope it will help someone.

Tested in Django 3.0.4.

In one line:

'Groupname' in user.groups.values_list('name', flat=True)

This evaluates to either True or False.

I have done it the following way. Seems inefficient but I had no other way in my mind:

@login_required
def list_track(request):

usergroup = request.user.groups.values_list('name', flat=True).first()
if usergroup in 'appAdmin':
    tracks = QuestionTrack.objects.order_by('pk')
    return render(request, 'cmit/appadmin/list_track.html', {'tracks': tracks})

else:
    return HttpResponseRedirect('/cmit/loggedin')

User.objects.filter(username='tom', groups__name='admin').exists()

That query will inform you user : “tom” whether belong to group “admin ” or not

I did it like this. For group named Editor.

# views.py
def index(request):
    current_user_groups = request.user.groups.values_list("name", flat=True)
    context = {
        "is_editor": "Editor" in current_user_groups,
    }
    return render(request, "index.html", context)

template

# index.html
{% if is_editor %}
  <h1>Editor tools</h1>
{% endif %}

I want users on the site to be able to download files whose paths are obscured so they cannot be directly downloaded.

For instance, I’d like the URL to be something like this: http://example.com/download/?f=somefile.txt

And on the server, I know that all downloadable files reside in the folder /home/user/files/.

Is there a way to make Django serve that file for download as opposed to trying to find a URL and View to display it?

For the “best of both worlds” you could combine S.Lott’s solution with the xsendfile module: django generates the path to the file (or the file itself), but the actual file serving is handled by Apache/Lighttpd. Once you’ve set up mod_xsendfile, integrating with your view takes a few lines of code:

from django.utils.encoding import smart_str

response = HttpResponse(mimetype='application/force-download') # mimetype is replaced by content_type for django 1.7
response['Content-Disposition'] = 'attachment; filename=%s' % smart_str(file_name)
response['X-Sendfile'] = smart_str(path_to_file)
# It's usually a good idea to set the 'Content-Length' header too.
# You can also set any other required headers: Cache-Control, etc.
return response

Of course, this will only work if you have control over your server, or your hosting company has mod_xsendfile already set up.

EDIT:

mimetype is replaced by content_type for django 1.7

response = HttpResponse(content_type='application/force-download')  

EDIT: For nginx check this, it uses X-Accel-Redirect instead of apache X-Sendfile header.

A “download” is simply an HTTP header change.

See http://docs.djangoproject.com/en/dev/ref/request-response/#telling-the-browser-to-treat-the-response-as-a-file-attachment for how to respond with a download.

You only need one URL definition for "/download".

The request’s GET or POST dictionary will have the "f=somefile.txt" information.

Your view function will simply merge the base path with the “f” value, open the file, create and return a response object. It should be less than 12 lines of code.

For a very simple but not efficient or scalable solution, you can just use the built in django serve view. This is excellent for quick prototypes or one-off work, but as has been mentioned throughout this question, you should use something like apache or nginx in production.

from django.views.static import serve
filepath = '/some/path/to/local/file.txt'
return serve(request, os.path.basename(filepath), os.path.dirname(filepath))

S.Lott has the “good”/simple solution, and elo80ka has the “best”/efficient solution. Here is a “better”/middle solution – no server setup, but more efficient for large files than the naive fix:

http://djangosnippets.org/snippets/365/

Basically, Django still handles serving the file but does not load the whole thing into memory at once. This allows your server to (slowly) serve a big file without ramping up the memory usage.

Again, S.Lott’s X-SendFile is still better for larger files. But if you can’t or don’t want to bother with that, then this middle solution will gain you better efficiency without the hassle.

Tried @Rocketmonkeys solution but downloaded files were being stored as *.bin and given random names. That’s not fine of course. Adding another line from @elo80ka solved the problem.
Here is the code I’m using now:

from wsgiref.util import FileWrapper
from django.http import HttpResponse

filename = "/home/stackoverflow-addict/private-folder(not-porn)/image.jpg"
wrapper = FileWrapper(file(filename))
response = HttpResponse(wrapper, content_type='text/plain')
response['Content-Disposition'] = 'attachment; filename=%s' % os.path.basename(filename)
response['Content-Length'] = os.path.getsize(filename)
return response

You can now store files in a private directory (not inside /media nor /public_html) and expose them via django to certain users or under certain circumstances.
Hope it helps.

Thanks to @elo80ka, @S.Lott and @Rocketmonkeys for the answers, got the perfect solution combining all of them =)

Just mentioning the FileResponse object available in Django 1.10

Edit: Just ran into my own answer while searching for an easy way to stream files via Django, so here is a more complete example (to future me). It assumes that the FileField name is imported_file

views.py

from django.views.generic.detail import DetailView   
from django.http import FileResponse
class BaseFileDownloadView(DetailView):
  def get(self, request, *args, **kwargs):
    filename=self.kwargs.get('filename', None)
    if filename is None:
      raise ValueError("Found empty filename")
    some_file = self.model.objects.get(imported_file=filename)
    response = FileResponse(some_file.imported_file, content_type="text/csv")
    # https://docs.djangoproject.com/en/1.11/howto/outputting-csv/#streaming-large-csv-files
    response['Content-Disposition'] = 'attachment; filename="%s"'%filename
    return response

class SomeFileDownloadView(BaseFileDownloadView):
    model = SomeModel

urls.py

...
url(r'^somefile/(?P<filename>[-\w_\\-\\.]+)$', views.SomeFileDownloadView.as_view(), name='somefile-download'),
...

It was mentioned above that the mod_xsendfile method does not allow for non-ASCII characters in filenames.

For this reason, I have a patch available for mod_xsendfile that will allow any file to be sent, as long as the name is url encoded, and the additional header:

X-SendFile-Encoding: url

Is sent as well.

http://ben.timby.com/?p=149

Try: https://pypi.python.org/pypi/django-sendfile/

“Abstraction to offload file uploads to web-server (e.g. Apache with mod_xsendfile) once Django has checked permissions etc.”

You should use sendfile apis given by popular servers like apache or nginx in production. Many years i was using sendfile api of these servers for protecting files. Then created a simple middleware based django app for this purpose suitable for both development & production purpose.You can access the source code here.
UPDATE: in new version python provider uses django FileResponse if available and also adds support for many server implementations from lighthttp, caddy to hiawatha

Usage

pip install django-fileprovider

• add fileprovider app to INSTALLED_APPS settings,
• add fileprovider.middleware.FileProviderMiddleware to MIDDLEWARE_CLASSES settings
• set FILEPROVIDER_NAME settings to nginx or apache in production, by default it is python for development purpose.

in your classbased or function views set response header X-File value to absolute path to the file. For example,

def hello(request):  
   // code to check or protect the file from unauthorized access
   response = HttpResponse()  
   response['X-File'] = '/absolute/path/to/file'  
   return response  

django-fileprovider impemented in a way that your code will need only minimum modification.

Nginx configuration

To protect file from direct access you can set the configuration as

 location /files/ {
  internal;
  root   /home/sideffect0/secret_files/;
 }

Here nginx sets a location url /files/ only access internaly, if you are using above configuration you can set X-File as,

response['X-File'] = '/files/filename.extension' 

By doing this with nginx configuration, the file will be protected & also you can control the file from django views

Django recommend that you use another server to serve static media (another server running on the same machine is fine.) They recommend the use of such servers as lighttp.

This is very simple to set up. However. if ‘somefile.txt’ is generated on request (content is dynamic) then you may want django to serve it.

Django Docs – Static Files

def qrcodesave(request): 
    import urllib2;   
    url ="http://chart.apis.google.com/chart?cht=qr&chs=300x300&chl=s&chld=H|0"; 
    opener = urllib2.urlopen(url);  
    content_type = "application/octet-stream"
    response = HttpResponse(opener.read(), content_type=content_type)
    response["Content-Disposition"]= "attachment; filename=aktel.png"
    return response 

Another project to have a look at: http://readthedocs.org/docs/django-private-files/en/latest/usage.html Looks promissing, haven’t tested it myself yet tho.

Basically the project abstracts the mod_xsendfile configuration and allows you to do things like:

from django.db import models
from django.contrib.auth.models import User
from private_files import PrivateFileField

def is_owner(request, instance):
    return (not request.user.is_anonymous()) and request.user.is_authenticated and
                   instance.owner.pk = request.user.pk

class FileSubmission(models.Model):
    description = models.CharField("description", max_length = 200)
        owner = models.ForeignKey(User)
    uploaded_file = PrivateFileField("file", upload_to = 'uploads', condition = is_owner)

I have faced the same problem more then once and so implemented using xsendfile module and auth view decorators the django-filelibrary. Feel free to use it as inspiration for your own solution.

https://github.com/danielsokolowski/django-filelibrary

Providing protected access to static html folder using https://github.com/johnsensible/django-sendfile: https://gist.github.com/iutinvg/9907731

I did a project on this. You can look at my github repo:

https://github.com/nishant-boro/django-rest-framework-download-expert

This module provides a simple way to serve files for download in django rest framework using Apache module Xsendfile. It also has an additional feature of serving downloads only to users belonging to a particular group