The situation is the following:
I have the url /app/categories/ that supports filtering by query arguments
/app/categories/ returns all categories
/app/categories/?project=1 returns all categories for the project with ID 1.
I want to also have an URL /app/projects/1/categories that will return the same result as /app/categories/?project=1 but without having to rewrite the view. Is it possible to make some kind of internal redirect or url rewriting such that when requesting /app/projects/1/categories the result will be the same as calling /app/categories/?project=1, but without redirecting? (in the future I might need to make the same thing for unsafe methods)
Make a common function that takes project id as argument and returns the categories object.
def get_categories(pk):
categories = Category.objects.filter(project=pk)
return categories
To add your url add this in urls.py,
url(r'^projects/(?P<pk>\d+)/categories/$', views.category_view,name="yourname"),
your views.py should look like this,
# /app/categories/?project=1
def category_view(request):
get_categories = get_categories(request.GET.get('project_id'))
# your logic
# /app/projects/1/categories
def other_category_view(request,pk):
get_categories = get_categories(pk)
# your logic
Here is my url :
url(r'^test/$|test/(\d+)', views.test_page)
So with the django runserver fired I can enter 127.0.0.1:8000/test/ or
the same url followed with a "page" number.
Here is a simplified version of my view :
def test_page(request, pagenumber):
paginator = Paginator(Test.objects.all(), 5)
page = 1
if pagenumber:
page = pagenumber
posts = paginator.page(page)
That works but is kinda inefficient. So I modified it to :
def test_page(request, page=1):
paginator = Paginator(Test.objects.all(), 5)
posts = paginator.page(page)
Which is nicer, works when I specify a page number in the url but when
I just enter 127.0.0.1:8000/test/ it doesn't. I got a :
Exception Type: TypeError
Exception Value: int() argument must be a string or a number,
not 'NoneType'
Exception Location: /usr/lib/python2.7/site-packages/django/core/
paginator.py in validate_number, line 23
why doesn't the attribute page take the default value 1 when I don't
specify any page number ?
For things like page numbers, it's better to use GET parameters, ie the form /test/?page=1. You do this directly in the view via request.GET.get('page'), so the urlconf is just r'^test/$.
Do a check to see if the function parameter page is None in the code. If it is set the value to be 1. Instead of relying on the function to default it.
I have never seen a url entry written that way.
Could you try something like:
To pass a named argument I think you have to specify that arguments name in your urls.py , it doesn't look like you're doing that now.
(?P<page>\d+)/$ this is the way django recommends to pass defualt arguments. https://docs.djangoproject.com/en/dev/topics/http/urls/#notes-on-capturing-text-in-urls
I would try this:
def test_page(request, **kwargs):
page = kwargs.get('pagenumber', 1)
paginator = Paginator(Test.objects.all(), 5)
posts = paginator.page(page)
No shell around so I couldn't test this code.
Please do not try to assign a default value to the argument "page" which you are passing in the function call. Just keep it as:
def test_page(request, page):
Please do try it, and write the url as:
url(r'^test/(?P<page>\d+)/$', views.test_page)
I'm just learning Django, and am getting stuck with some url logic. I'm trying to allow either a category name or id in the url:
...
url(r'^(?P<booze_q>\w+|\d+)/$','glasses.views.booze'),
...
And then in thew view, only deal with that result once. However, if the url is a string - in this case, Whiskey, I get an error for trying to pass a string where an int is expected. This is the closest I've gotten so far:
def booze(request, booze_q):
booze = get_object_or_404(Booze,Q(pk=booze_q)|Q(name=booze_q))
return render_to_response('booze/detail.html', {'booze': booze})
But this returns an error: invalid literal for int() with base 10: 'Whiskey'
I'm sure it's a pretty easy thing, but this is my first Django app, so any help would be appreciated.
tl;dr: End result, I'd like mysite.com/1/ or mysite.com/Whiskey/ to both call the glasses.views.booze view, and get the object with id=1 or name=Whiskey
This is a common scenario you'll encounter quite often, which is typically handled by resorting to multiple arguments and having views behave differently based on which of the view arguments are then present or not.
What you do is first define a URL pattern that uniquely matches each specific case and then let Django's URL resolver set the arguments accordingly based on which of the patterns was matched.
Here's an example with a class based view, that performs two different queries based on which of the two keyword arguments, booze_id or booze_name, is set:
url(r'^(?P<booze_id>\d+)/$', BoozeDetailView.as_view()),
url(r'^(?P<booze_name>\w+)/$', BoozeDetailView.as_view()),
class BoozeDetailView(DetailView):
model = Booze
def get_object(self):
booze_id = self.kwargs.get('booze_id', None)
booze_name = self.kwargs.get('booze_name', None)
if booze_id:
return self.model.objects.get(id=booze_id)
else:
return self.model.objects.get(name=booze_name)
You will always get a string, even if the string contains a number.
1) You should not have a parameter that could be either an id or something else. One day you will enter an item whose name is a number and your app will fail.
2) When querying for pk with a string django automatically tries to convert it into an integer. You'll have to handle the non-pk case before constructing that query.
I want the WHOLE url after the slash to be passed to a script. If I do :
url(r'^(?P<id>.*)$', alias.get, name="alias"),
Then I only get the path component and not the query parameters passed to my function. I then have to :
def urlencode(dict) :
if len(dict) == 0 : return ""
params = {}
for k, v in dict.items() :
params[k] = v.encode('utf-8')
return "?" + urllib.urlencode(params)
def get(id) :
id += urlencode(request.GET)
I am doing this for a lot of my views and I keep forgetting it and creating bugs. Is there any way to tell my urls.py to match everything including the query string?
Inside your view, you can get the URL including the query string with request.get_full_path().
There's also request.build_absolute_uri() if you want to include the http://servername.com part.
No, there's no way of doing that. The GET parameters aren't passed to urls.py.
I wonder why you need to do this though. Why are so many of your views dependent on GET querystrings? Why don't you use the Django way of doing things, which is to make the parameters part of the URL itself, rather than as querystrings.
I am currently defining regular expressions in order to capture parameters in a URL, as described in the tutorial. How do I access parameters from the URL as part the HttpRequest object?
My HttpRequest.GET currently returns an empty QueryDict object.
I'd like to learn how to do this without a library, so I can get to know Django better.
When a URL is like domain/search/?q=haha, you would use request.GET.get('q', '').
q is the parameter you want, and '' is the default value if q isn't found.
However, if you are instead just configuring your URLconf**, then your captures from the regex are passed to the function as arguments (or named arguments).
Such as:
(r'^user/(?P<username>\w{0,50})/$', 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."