how to create params in django url? - django

I have a django project where url is like this
url(r'^invoice/(?P<invoice_id>[A-Za-z0-9]+)/(?P<order_id>[A-Za-z0-9]+)$',GenerateInvoicePdf,name='invoice'),
which generates url localhost:8000/invoice/2341wq23fewfe1231/3242
but i want url to be like localhost:8000/invoice?invoice_id=2341wq23fewfe1231&order_id=3242
i tried documentation and used syntax like this re_path(r'^comments/(?:page-(?P<page_number>\d+)/)?$', comments), But did not get desired result.
how can i do so?

The parts which you are trying to write after ? is called url query string. You don't need to define them in the urls.py. You can just use:
re_path(r'^comments/$', comments),
And inside comments views, you can access the query string like this:
def comments(request):
invoice_id = request.GET.get('invoice_id')
order_id = request.GET.get('order_id')
# rest of the code

Related

How to recognize urls with get parameters un django?

I am learning Django, usually I use urls like www.mysite.com/post/1, and I have understood them because they are well described in the documentation.
Now I need to create a url like: www.mysite.com/calendar/?day=21&?month=12&year=2020, I don't understand what should I write in my ulrpatterns list.
I have tried something like this:
url(r'^search/(?P<parameter>\w+)', views.calendar, name='calendar')
but it doesn't work.
What should I write as regex expression?
Thanks
These parameters are not part of the path, you can't capture them using the url patterns.
You can access them directly inside your view by using request.GET, see the docs.
A common pattern is the following:
def calendar(request):
day = request.GET.get("day") # Will be None if "day" isn't in the query
month = request.GET.get("month")
year = request.GET.get("year")
[...]

django: access url name pattern in cbv

I would like to map several urls to the same view.
url(r'^foo/', BaseView.as_view(), name='foo'),
url(r'^bar/', BaseView.as_view(), name='bar'),
url(r'^buzz/', BaseView.as_view(), name='buzz'),
Is there anyway in my class based view I can access the name parameters? I know it's possible to specify it in as_view(). But is there a way around that extra boilerplate code?
What about:
url(r'^/(?P<name>[-\w]+)/$', BaseView.as_view())
Then you can just grab the name from BaseView:
name = self.kwargs['name']

Django request.GET.get() always returns "None"

I have a URL: http://localhost:8000/submit_workout/2/
This URL was created with: r'^submit_workout/(?P<wr_id>\d+)/$'
I am trying to retrieve "wr_id" when a form on that page is submitted.
I'm trying: wr_id = request.GET.get('wr_id',None) and am expecting wr_id=2 but keep getting wr_id=None returned.
Any thoughts? I am new to programming/django and really appreciate your time and expertise.
URL params, which are named in the url regex, can be passed as arguments to your method which handles the request. If your dispatcher looks like this:
urlpatterns = patterns('',
(r'^submit_workout/(?P<wr_id>\d+)/$', 'submit_workout'),
Then your method should look like this:
def submit_workout(request, wr_id):
and wr_id can be accessed directly.
If you want wr_id to be a GET variable, then your url should look like this:
http://localhost:8000/submit_workout?wr_id=2

Django query: field is substring

In Django I have my Site model that contains the field "base_url" that's the base url of the site. I've an object like this:
foo = Site(base_url="http://foo_base_url.com/")
foo.save()
I receive an url and I want to obtain the site object having this url. I would like to perform a query in django like this:
Site.objects.get(base_url__is_substring="http://foo_base_url.com/something_non_base_url")
How can I perform this query?
Thanx
edit:
It doesn't exist a pattern for base_url, my foo site can be:
foo = Site(base_url="http://foo.com/base/url/")
What you want is not provided by the Django ORM but you can use the where param described under the reference for QuerySet:
url = "http://foo_base_url.com/something_non_base_url"
Site.objects.extra(where=["%s LIKE CONCAT('%%',field,'%%')"], params=[url]).get()
Keep in mind that there is no standard method of concatenation across DMBS so if you migrate, you'll have to migrate this code.
The only portable method would be to filter it using Python:
sites = [site for site in Site.objects.all() if site.base_url in url]
although this is of course not ideal for huge data sets.

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.
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."