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Django: Passing parameters in URL as query arguments

How can I pass parameters via URL as query parameters to avoid multiple and complicated url patterns?
For example, instead of making a complicated url like
example.com/page/12/red/dog/japan/spot
or something like that, and then a corresponding entry in urls.py that will parse that url and direct it to a view, I want to simply get a url where I can freely add or remove parameters as needed similar to the ugly way
example.com/page?id=12&color=red&animal=dog&country=Japan&name=spot
Then in urls.py simply have something like
path('page/<parameter_dictionary>', views.page, name='page' parameters='parameter_dictionary)
If I have to use url patterns, how can I account for urls that have parameters that may or may not fit the pattern, such as sometimes
"/page/12/red/dog/Japan/spot" -> path('page/<int:id>/<str:color>/<str:animal>/<str:country>/<str:name>', views.page, name='page'),
"/page/12/dog/red/Japan/"-> path('page/<int:id>/<str:animal>/<str:color>/<str:country>', views.page, name='page')
"/page/dog/red/Japan/"-> path('page/<str:animal>/<str:color>/<str:country>', views.page, name='page')
I would like to just have anything sent to http://example.com/page/
go to views.page(), and then be accessible by something like
animal = request.GET['animal']
color = request.GET['color']
id = request.GET['id']
etc. so examples below would all work via one entry in urls.py
example.com/page?id=12&animal=dog&country=Japan&name=spot
example.com/page?id=12&color=red&animal=dog&name=spot
example.com/page?id=12&country=Japan&color=red&animal=dog&name=spot

You are looking for queryparameters and you are almost done with it. The following code is untested but should kinda work:
def page(request):
animal = request.GET.get("animal",None) # default None if not present
color = request.GET.get("color",None)
return render(request,'some_html.html')
# urls.py:
path('page/', views.page, name='page')
You access the queryparameters via the passed request object request.GET. This is a dict like object. The main difference is that this object handles multi keys.
For example if you pass the these params ?a=1&a=2 to your url, it converts request.GET.getlist("a") # Returns ["1","2"] to a list.
request.GET.get("a") returns the last passed value "2" as #Kbeen mentioned in comments,. Read more about QueryDict here.
Also be sure to know the difference and best practice for url parameters and queryparameters. Example Stackoverflow post
Edit: Added request.GET.getlist()

Django - Testing a view name retrieved from a url parameter ie. ref=home

Simply put, I am trying to test a view name, captured from a URL parameter, within one of my view functions to see if it is valid. If not, to redirect to another page.
For example, take this url...
www.site.com/users/?ref=view_name
Then in my view
ref = request.GET.get('ref', None)
return redirect(ref) if ref else redirect('users')
The problem is, of course, that if a user alters the ref parameter, it will kick back a 404. I was hoping to be able to test if ref=home is a valid view and return it, if not then redirect to another view.
I have been messing with Django's resolve and reverse but I am not getting the results I was looking for.
I also tried to use a try/finally in all sorts of ways mixed in with resolve and reverse. Yeah, dumb...I know.
try:
if ref:
return redirect(resolve(ref))
finally:
return redirect('user')
I have searched for almost two hours to try to fund a succinct way to do this.
Any help would be appreciated!

The problem is that you are passing a view name to resolve() which requires a URL path. To protect against 404, you need to first reverse() the view name to get a path. Then you can use that path with resolve() to check if the path exists.

Working Solution
This was actually very simple after guidance from #Code-Apprentice
A couple Imports first off
from django.urls.resolvers import NoReverseMatch
from django.urls import reverse
Then to get the ref parameter and validate is as a legit view name I did the following
# get ref parameter
ref = request.GET.get('ref', None)
if ref:
try:
# redirect to ref view name if valid
return redirect(reverse(ref))
except NoReverseMatch as e:
print('No reverse match found')
# redirect to specific view if ref is invalid
return redirect('users')

Get all parameters and their values in a request in Django [duplicate]

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.

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 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.

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)
]

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

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("&")

url parameters may be captured by request.query_params

It seems more recommended to use request.query_params. For example,
When a URL is like domain/search/?q=haha, you would use request.query_params.get('q', None)
https://www.django-rest-framework.org/api-guide/requests/
"request.query_params is a more correctly named synonym for request.GET.
For clarity inside your code, we recommend using request.query_params instead of the Django's standard request.GET. Doing so will help keep your codebase more correct and obvious - any HTTP method type may include query parameters, not just GET requests."

where to define a slug without a model.py in Django

I am rather new to django, and am looking at where to define a slug in django when creating a backend without models. the url is created as such:
url(r'^main/(?P<slug>[-\w]+)/', include('main.urls')),
I have slugs within my main.urls which I define inside of each view function. Im not exactly sure where to define this slug(link, whatever you may call it). On other django slug examples, the common way is in a model, and I am currently talking to a program rather then creating my own models.
Would this be in the urls.py, or views.py (in the project, not app)?
Thank you so much. Hopefully this is understandable.

It's not hard. Really.
In url-configs each entry is simply a regular expression which has to match a url that is visited by an end user. r'^main/(?P<slug>[-\w]+)/' will for example match with: http://localhost:8000/main/some-slug/
You can use a special kind of syntax in your regular expression to extract matched data and pass that data as a variable to your view function.
The bit that does that is (?P<slug>[-\w]+) it puts matched words (in this case a slug) into a variable called slug (the <slug> part, it defines the variable name). In this humble example the slug variable will be set to "some-slug".
The variable will be accessible in your view like this:
from django.http import HttpResponse
def handle_my_view(request, slug=homepage):
# do stuff with slug
return HttpResponse("I did stuff with slug: {}".format(slug))
Learn more about, and fiddle with regular expressions
At http://www.regexr.com
But why do i see slugs used in models?:
A slug (or named variable, coming from a url 'interception') can be used for anything. Commonly the slug variable itself will be used to retrieve a database record of some sorts... And that involves using models.
You can do whatever you want with them; add stuff, subtract stuff, capitalize, whatever. The sky is the limit.
From the Django docs:
https://docs.djangoproject.com/en/1.10/topics/http/urls/#named-groups
Named groups
The above example used simple, non-named regular-expression groups (via parenthesis) to capture bits of the URL and pass them as positional arguments to a view. In more advanced usage, it’s possible to use named regular-expression groups to capture URL bits and pass them as keyword arguments to a view.
In Python regular expressions, the syntax for named regular-expression groups is (?Ppattern), where name is the name of the group and pattern is some pattern to match.
Here’s the above example URLconf, rewritten to use named groups:
from django.conf.urls import url
from . import views
urlpatterns = [
url(r'^articles/2003/$', views.special_case_2003),
url(r'^articles/(?P<year>[0-9]{4})/$', views.year_archive),
url(r'^articles/(?P<year>[0-9]{4})/(?P<month>[0-9]{2})/$', views.month_archive),
url(r'^articles/(?P<year>[0-9]{4})/(?P<month>[0-9]{2})/(?P<day>[0-9]{2})/$', views.article_detail),
]
This accomplishes exactly the same thing as the previous example, with one subtle difference: The captured values are passed to view functions as keyword arguments rather than positional arguments. For example:
A request to /articles/2005/03/ would call the function views.month_archive(request, year='2005', month='03'), instead of views.month_archive(request, '2005', '03').
A request to /articles/2003/03/03/ would call the function views.article_detail(request, year='2003', month='03', day='03').
In practice, this means your URLconfs are slightly more explicit and less prone to argument-order bugs – and you can reorder the arguments in your views’ function definitions. Of course, these benefits come at the cost of brevity; some developers find the named-group syntax ugly and too verbose.

Django - get names of parameters needed to reverse url

Background
Let's say I have a url pattern with parameters that will link me to a view in django:
url(
r'^things/(?P<thing_name>\w+)/features/(?P<feature_name>\w+)$',
views.thingFeature,
name='thing_feature'
),
And lets say I have a thing and a feature:
thing = Thing.objects.get(.....)
feature = thing.feature_set.first()
t_name = thing.name
f_name = feature.name
Now Django gives me the awesome ability to get a url that brings me to a page dedicated to a specific feature of a specific thing. I can do that like so:
from django.core.urlresolvers import reverse
url = reverse('thing_feature', thing_name=t_name, feature_name=f_name)
# url == '/things/thing2/features/left-arm'
Question
Now I've stumbled into a situation that I need to specifically address. I'm not looking for a workaround - I'm looking to solve the following problem:
Given a url's name, how do I get the list of kwarg argument names needed to reverse that url?
I am looking for the function get_kwarg_names_for_url. It behaves like so:
url_kwarg_names = get_kwarg_names_for_url('thing_feature')
# url_kwarg_names == ['thing_name', 'feature_name']
url_kwarg_names is now the list of every keyword I need to supply to Django's reverse function in order to reverse the url named "thing_feature".
Any help is appreciated!
Solution
Based on knbk's answer I was able to come up with the following solution:
def get_kwarg_names_for_url(url_name):
resolver = get_resolver(get_urlconf())
reverse_data = resolver.reverse_dict[url_name]
pattern_list = reverse_data[0]
'''
Need to specify the 1st pattern because url regexes can
potentially have multiple kwarg arrangments - this function does
not take this possibility into account.
'''
first_pattern = pattern_list[0]
'''
`first_pattern` is now of the form `(url_string, kwarg_list)` -
all we are interested in is the 2nd value.
'''
return first_pattern[1]

I'll start with a fair warning: this isn't possible using the public API. On top of that, I'm actively working to rewrite the URL dispatcher for 1.10, so this method will most likely break by that time.
First, you need to get the right RegexURLResolver. If the view is not in a namespace, you can use the reverse_dict to get a list of possibilities, and extract the kwargs:
def get_kwargs(view_name):
resolver = urlresolvers.get_resolver()
patterns = resolver.reverse_dict.getlist(view_name)
kwargs = []
for possibility, pattern, defaults in patterns:
for result, params in possibility:
kwargs.append(params)
return kwargs
Since a view name can have multiple patterns with different kwargs (though you'd want to avoid that for your own sanity), this will return a list of each set of possible kwargs. Usually the different sets would be the required kwargs on one side and required + optional kwargs on the other side.
I haven't tested this, but if it doesn't work you can dig around in resolver.reverse_dict for a bit to find out the exact specifics. It wasn't exactly designed with usability in mind.

You should be able to do this with a resolve()
From the Docs:
A ResolverMatch object can then be interrogated to provide information about the URL pattern that matches a URL:
func, args, kwargs = resolve('/some/path/')
Specific to your example code:
url = reverse('thing_feature')
func, args, kwargs = resolve(url)
# args == ['thing_name', 'feature_name']