Django: Make query using url ( Assigning variable using url )

Question

I am trying to make a query system for my website, i think the best way and the most compact would be to assign search variable using url pattern.

So for example, i want to search objects of model User:

User sends HttpRequest to following url:

https://127.0.0.1/search/q="admin"

Now HttpRequest is also sent to search view, we somehow get q variable data.

def search(request):
    for query in User.objects.all():
        if q in query: # < We somehow need to get data of 'q'.
           return HttpResponse(q)

Since i have admin in User.objects.all(), this should return HttpResponse of 'admin'.

How can this url pattern be made? So i can assign q variable from the url and then send it to system to find it?

You can use query parameters, but for this your url should look like the following: https://127.0.0.1/search/?q=admin (note the question mark). Then, in your view, you can access all of your query parameters using request.GET[param] (in your case request.GET['q']). You can read this wiki article to find out more about query parameters. — iulian
– iulian, Commented Nov 7, 2016 at 12:21
Did you have a look at this question already? stackoverflow.com/questions/150505/… The official documentation on how to handle GET-parameters in Django can be found here: docs.djangoproject.com/en/1.10/ref/request-response/… — maennel
– maennel, Commented Nov 7, 2016 at 12:22

RemcoGerlich · Accepted Answer · 2016-11-07 12:53:33Z

1

I have problems with this URL:

https://127.0.0.1/search/q="admin"

There is no ? in the URL, so there is no query string, it's all part of the "path". Using characters like = and " in there will confuse a lot of things, if it works at all.

Either just do

https://127.0.0.1/search/admin

With an URL pattern like r'^search/(?P<querystring>.+)$', or

https://127.0.0.1/search/?q=admin

In this case the query string will be in request.GET['q']; it's also possible to use Django forms to process query parameters (e.g. for validating them).

answered Nov 7, 2016 at 12:53

RemcoGerlich

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1 Comment

ShellRox Over a year ago

I knew the first method, but didn't find it as good design, second method is what i needed.

Lauri Elias · Accepted Answer · 2016-11-07 12:47:33Z

1

You can capture named strings from URLs like this:

urls.py:

urlpatterns = [
    url(r'^blog/page(?P<num>[0-9]+)/$', views.page),
]

views.py:

def page(request, num="1"):

answered Nov 7, 2016 at 12:47

Lauri Elias

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