I am passing a variable in my URL:
mydomain.com/app/?next_page=my_page
I can get this variable in a view with:
def mypage(request):
var = request.GET['next_page']
Is it best practice to also change the URL to require next_page? Something along the lines of:
path('app/?nextpage=<str>', mypage, name='my_page')
What's best practice? If so, what's the correct syntax for this (I know the example is incorrect)?
It depends on your needs.
Do not define a fixed url route; if you use the query parameters for filtering and there is more than one possible parameter
Example: "app/photos?size=100x100" and "app/photos/?color=blue"
Define a fixed url route; if it will be the same for each and every page, like details of a particular page:
Example: "app/orders/123123123" and "app/orders/123123123"
Btw, the correct syntax is:
path(app/<str:next_page>/, mypage, name="my_page")
You should take a look at path patterns. Enforcing a GET parameter in a path is not really a good practice. So if you want to require a username for example you can use:
path('bio/<username>/', views.bio, name='bio'),
You can find more patterns in Django documentation to catch strings, slugs, integers etc.
And in views you should define your function as such:
def mypage(request, username):
...code...
About GET:
Keep in mind that request.GET["value"] will raise a ValueError if that parameter does not exist. So you can catch that error to inform user that they are missing a parameter. (This will make this parameter obligatory.)
You can also use request.GET.get("value") which will return None if the key does not exist. If you want to use a default parameter you can use of course, request.GET.get("value", "default")
You can use as many parameters as you want in your link with or without path patterns. Their values will be stored in request.GET
I have some code that is repeated at the start of my Django views. It basically just adds some variables to the context, but based on the URL argument, e.g.
def someView(request, id):
target = Target.objects.get(id=id)
# name will be added to ctx
name = target.name
(there are more attributes added and other attributes from related models, but this gives the general idea --- There are quite a few lines of repeat code at the start of each view)
I thought I could make my code more DRY by taking advantage of Django's context processors, but it would seem these don't access to the URL arguments?
Is there another way to avoid these repeat lines? Maybe middleware or something else?
You can access the URL parameters via request through the resolver_match attribute. So for instance you can do request.resolver_match.kwargs['id'] to get the ID kwarg.
I have in a Django project all my urls based on the following syntax:
/ID_PROGRAM/ID_PROJECT/blablabla
I would like by default that all my queries have the following filters:
.filter(program=ID_PROGRAM).filter(project=ID_PROJECT)
How can I apply these filters automatically to all my queries? My idea was to define a new manager. But is the manager able to access to the url parameters? I this the best way to do?
To complet the question, I want to enrich all my queries without having to pass explicitly the view parameters to the manager.
You could have just tried it to see if it works.
Yes, managers do accept parameters
class MyModelManager(models.Manager):
def my_filters(self, id_prog, id_proj):
return super(MyModelManager, self).get_query_set().filter(program=id_prog, project=id_proj)
and in the views:
MyModelManager.objects.my_filters(id_prog, id_proj)
Documentation on custom managers
Python promotes "Explicit is better than implicit"
karthikr is almost right, but you can also use:
1 - decorator above your function. Decorator will get args from url and put objects to any variable
2 - write mixin and aply it to view. Mixin will get args from url at overriden dispatch and save filter result to self.custom_context. Override get_context_data to merge contexts.
my url pattern looks like that:
(r'^fb/custom/(?P[a-zA-Z0-9+]*)/admin/', include(custom_admin_site.urls)),
I overrode the admin_view methode of my admin site:
def admin_view(self, view, cacheable=False):
def inner(request, *args, **kwargs):
if kwargs.has_key('custom_id'):
request.custom_id = kwargs.pop('custom_id')
return view(request, *args, **kwargs)
if not cacheable:
inner = never_cache(inner)
# We add csrf_protect here so this function can be used as a utility
# function for any view, without having to repeat 'csrf_protect'.
if not getattr(view, 'csrf_exempt', False):
inner = csrf_protect(inner)
return update_wrapper(inner, view)
This way I don't need the paramter custom_id in the view methods like index. My problem is that i can't use urlresolvers.reverse('custom-admin:index').
If I use it without parameter i get this error:
Page not found. Request URL: http://localhost:8000/fb/custom/(?P%3Ccustom_id%3E[a-zA-Z0-9%5C+]*)/admin/
Ok no suprise. I didn't provide the parameter custom_id. But with the paramter I get this error:
reverse() got an unexpected keyword argument 'custom_id'
Any idea how to solve this. I would really like to use the reverse lookup. The url template tag has the same problem.
A few issues with your url pattern:
Unless your are trying to match empty string for your custom id (which I would guess you are not) you should use + instead of *.
The + sign should go outside of your brackets.
If you are just trying to match letters and numbers, your regex can be simplified to [\w]+
Finally, if you want a named argument for custom id, you must include the name in your pattern.
So ultimately, your url patter should look like this:
(r'^fb/custom/(?P<custom_id>[\w]+)/admin/', include(custom_admin_site.urls)),
Now I'm not sure how you're trying to call urlresolvers.reverse, but if you need to pass args or kwargs, it should something look like one of these:
urlresolvers.reverse('custom-admin:index', args=[custom_id])
or for kwargs, such as in the pattern I included above:
urlresolvers.reverse('custom-admin:index', kwargs={'custom_id':custom_id})
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."