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()
I cannot for the life of me find a way to redirect to another view with arguments for that view in Django. I've tried the django.shortcuts redirect method, but that doesn't seem to work (it seems to want to take patterns for the url pattern and not the view function. The HttpResponseRedirect function obviously doesn't work because that function just redirects to another url. What other options are there?
Seems you want to give use redirect and want to give a parameter thats so easy you can do that
x.order_total= total+shipping_total
x.save()
return redirect(f'/order/payment/{order_number}')
here i have pasted just a snippet of my one view which is doing that you can insert the parameter using f outside the string if you have matching url it will work fine
You can redirect to other view with parameter(s) using reverse that doesn't use hard coded url instead the name of url infact redirect uses that internally
def view_one(request):
return redirect('view_two', arg=arg)
By default redirect will do temporary redirect for perminant redirect you have to specify it as to know the difference between the two take a look here
You can use the reverse function to fill your URL pattern
You can import the reverse function with the following line:
from django.core.urlresolvers import reverse
If you have URLs like this you can use reverse:
url(r'^project/(?P<project_id>\d+)/$','user_profile.views.EditProject',name='edit_project'),
path('project/<str:project_id>/', name='edit_project'),
usage:
redirect(reverse('edit_project', kwargs={'project_id':4}))
or:
redirect(reverse('app_name:edit_project', kwargs={'project_id':4}))
Doc here
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')
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."
how do i redirect a registered user to his/her db.table.id 'view' without going through smartgrid in web2py?
i have tried using:
redirect(URL(f='first', args=['mydata/view', 'mydata/%s', %request.vars.name]))
where mydata is the view for my table db.mydata and 'first' is my function.
It always returns to the smartgrid interface.
There are two problems. First, the final URL argument must be the record ID, but it looks like you are instead using a name (i.e., request.vars.name). Second, by default, the grid uses signed URLs, so you must either disable the signatures (not recommended) or add a user signature to the URL you generate. So, the link should be something like this:
redirect(URL(f='first', args=['mydata', 'view', 'mydata', request.vars.id],
user_signature=True))
Also, note that in the args list, each element can (and generally should) be a separate URL arg. So, instead of ['mydata/view', ...], it should be ['mydata', 'view', ...].