django username in url, instead of id

Question

in a mini virtual community, i have a profile_view function, so that i can view the profile of any registered user. The profile view function has as a parameter the id of th

Answer 1

Here's how I do it using class based generic views and the use of slugs. It's surprisingly easy and requires only a few lines of code with the help of slugs.

# accounts/views.py
from django.contrib.auth.models import User
from django.views.generic.detail import DetailView

class UserProfileView(DetailView):
    model = User
    slug_field = "username"
    template_name = "userprofile.html"

# accounts/urls.py
from views import UserProfileView
urlpatterns = patterns('',
    # By user ID
    url(r'^profile/id/(?P<pk>\d+)/$', UserProfileView.as_view()),
    # By username
    url(r'^profile/username/(?P<slug>[\w.@+-]+)/$', UserProfileView.as_view()),
)

Now you can access both the user like accounts/profile/id/123/ as well as accounts/profile/username/gertvdijk/.

What's happening here?

• The usually required pk URL parameter is omitted and replaced by slug in the URL pattern. This is accepted by the view, because...
• The slug parameter is being used by the DetailView (or any other SingleObjectMixin-based view) to find the object on User.username by the use of model = User and slug_field = "username". (docs)
• Because a slug field is optional, the view is also happily serving using just the pk parameter.

Answer 2

For Django 2.0 and above:

path(‘profile/<username>/, views.profile, name=‘profile’),

Answer 3

In urls.py

urlpatterns = [
    path('<str:username>/', UserDetailView.as_view(), name='user_detail'),
]

With class based view.

class UserDetailView(LoginRequiredMixin, DetailView):

    model = User
    slug_field = "username"
    slug_url_kwarg = "username"
    template_name = "account/user_detail.html"

    def get_object(self):

        object = get_object_or_404(User, username=self.kwargs.get("username"))

        # only owner can view his page
        if self.request.user.username == object.username:
            return object
        else:
            # redirect to 404 page
            print("you are not the owner!!")

I have tested in Django 2.1.

Answer 4

You don't show what you have tried, or where you are having trouble. This is fairly simple, after all - just pass a username string to the function instead of an integer, and look up based on that.

def profile_view(request, username):
    u = User.objects.get(username=username)

and modify your url to allow strings rather than integers:

url(r'^profile_view/(?P<username>\w+)/$', 
                       profile_view,
                       name='profile_view'),

Answer 5

Add in the top of the file:

from django.shortcuts import get_object_or_404

and replace

u = User.objects.get(pk=user)

with

u = get_object_or_404(User, <login>=user)

where login is the username field in User model.