Capturing URL parameters in request.GET

I am currently defining regular expressions in order to capture parameters in a URL, as described in the tutorial. How do I access parameters from the URL as part the HttpRequest object?

My HttpRequest.GET currently returns an empty QueryDict object.

I'd like to learn how to do this without a library, so I can get to know Django better.

When a URL is like domain/search/?q=haha, you would use request.GET.get('q', '').

q is the parameter you want, and '' is the default value if q isn't found.

However, if you are instead just configuring your URLconf**, then your captures from the regex are passed to the function as arguments (or named arguments).

Such as:

(r'^user/(?P<username>\w{0,50})/$', views.profile_page,),

Then in your views.py you would have

def profile_page(request, username):
    # Rest of the method

To clarify camflan's explanation, let's suppose you have

• the rule url(regex=r'^user/(?P<username>\w{1,50})/$', view='views.profile_page')
• an incoming request for http://domain/user/thaiyoshi/?message=Hi

The URL dispatcher rule will catch parts of the URL path (here "user/thaiyoshi/") and pass them to the view function along with the request object.

The query string (here message=Hi) is parsed and parameters are stored as a QueryDict in request.GET. No further matching or processing for HTTP GET parameters is done.

This view function would use both parts extracted from the URL path and a query parameter:

def profile_page(request, username=None):
    user = User.objects.get(username=username)
    message = request.GET.get('message')

As a side note, you'll find the request method (in this case "GET", and for submitted forms usually "POST") in request.method. In some cases, it's useful to check that it matches what you're expecting.

Update: When deciding whether to use the URL path or the query parameters for passing information, the following may help:

• use the URL path for uniquely identifying resources, e.g. /blog/post/15/ (not /blog/posts/?id=15)
• use query parameters for changing the way the resource is displayed, e.g. /blog/post/15/?show_comments=1 or /blog/posts/2008/?sort_by=date&direction=desc
• to make human-friendly URLs, avoid using ID numbers and use e.g. dates, categories, and/or slugs: /blog/post/2008/09/30/django-urls/

Using GET

request.GET["id"]

Using POST

request.POST["id"]

Someone would wonder how to set path in file urls.py, such as

domain/search/?q=CA

so that we could invoke query.

The fact is that it is not necessary to set such a route in file urls.py. You need to set just the route in urls.py:

urlpatterns = [
    path('domain/search/', views.CityListView.as_view()),
]

And when you input http://servername:port/domain/search/?q=CA. The query part '?q=CA' will be automatically reserved in the hash table which you can reference though

request.GET.get('q', None).

Here is an example (file views.py)

class CityListView(generics.ListAPIView):
    serializer_class = CityNameSerializer

    def get_queryset(self):
        if self.request.method == 'GET':
            queryset = City.objects.all()
            state_name = self.request.GET.get('q', None)
            if state_name is not None:
                queryset = queryset.filter(state__name=state_name)
            return queryset

In addition, when you write query string in the URL:

http://servername:port/domain/search/?q=CA

Do not wrap query string in quotes. For example,

http://servername:port/domain/search/?q="CA"

def some_view(request, *args, **kwargs):
    if kwargs.get('q', None):
        # Do something here ..

For situations where you only have the request object you can use request.parser_context['kwargs']['your_param']

You have two common ways to do that in case your URL looks like that:

https://domain/method/?a=x&b=y

Version 1:

If a specific key is mandatory you can use:

key_a = request.GET['a']

This will return a value of a if the key exists and an exception if not.

Version 2:

If your keys are optional:

request.GET.get('a')

You can try that without any argument and this will not crash. So you can wrap it with try: except: and return HttpResponseBadRequest() in example. This is a simple way to make your code less complex, without using special exceptions handling.

I would like to share a tip that may save you some time.
If you plan to use something like this in your urls.py file:

url(r'^(?P<username>\w+)/$', views.profile_page,),

Which basically means www.example.com/<username>. Be sure to place it at the end of your URL entries, because otherwise, it is prone to cause conflicts with the URL entries that follow below, i.e. accessing one of them will give you the nice error: User matching query does not exist.

I've just experienced it myself; hope it helps!

These queries are currently done in two ways. If you want to access the query parameters (GET) you can query the following:

http://myserver:port/resource/?status=1

request.query_params.get('status', None) => 1

If you want to access the parameters passed by POST, you need to access this way:

request.data.get('role', None)

Accessing the dictionary (QueryDict) with 'get()', you can set a default value. In the cases above, if 'status' or 'role' are not informed, the values ​​are None.

This is not exactly what you asked for, but this snippet is helpful for managing query_strings in templates.

If you only have access to the view object, then you can get the parameters defined in the URL path this way:

view.kwargs.get('url_param')

If you only have access to the request object, use the following:

request.resolver_match.kwargs.get('url_param')

Tested on Django 3.

If you don't know the name of params and want to work with them all, you can use request.GET.keys() or dict(request.GET) functions

This is another alternate solution that can be implemented:

In the URL configuration:

urlpatterns = [path('runreport/<str:queryparams>', views.get)]

In the views:

list2 = queryparams.split("&")

You might as well check request.META dictionary to access many useful things like PATH_INFO, QUERY_STRING

# for example
request.META['QUERY_STRING']

# or to avoid any exceptions provide a fallback

request.META.get('QUERY_STRING', False)

you said that it returns empty query dict

I think you need to tune your url to accept required or optional args or kwargs Django got you all the power you need with regrex like:

url(r'^project_config/(?P<product>\w+)/$', views.foo),

more about this at django-optional-url-parameters

views.py

from rest_framework.response import Response

def update_product(request, pk):
	return Response({"pk":pk})

pk means primary_key.

urls.py

from products.views import update_product
from django.urls import path

urlpatterns = [
    ...,
	path('update/products/<int:pk>', update_product)
]

url parameters may be captured by request.query_params