Take a url pattern as input in django

Question

say i have a variable :

a=['https://www.google.com','https://www.go.com','https://www.ogle.com']

i have a Template dome.html:

   {%for i in a %}

   <a href=/link/{{a}}>Click</a><br/>

   {%endfor%}

I have views.py as bellow:

 def someview(request,id):
    do something with input (id)

Now i want to configure my urls.py so that it will communicate with my views.py which take {{a}} as input (id).

url(r'^/link/(url-regex)',someview , name="someview"),

Now i actually want thet url-regex , how i am going to configure it . Kindly help

May be you could do, {% for i in a %} <a href="{{i}}">Click</a><br/> {% endfor %} — zaidfazil
– zaidfazil, Commented Jun 26, 2017 at 4:54
Its not clear, what you are implying.. Please explain it a little bit further. — zaidfazil
– zaidfazil, Commented Jun 26, 2017 at 4:56
But, {{ i }} is not an id, its just uri. Do you want to go to the uri? — zaidfazil
– zaidfazil, Commented Jun 26, 2017 at 5:00
i want u url pattern that will send {{i}} to the someview(request,id) fiunction . like url(r'^/link/(urlpattern)', someview, name="someview"), — anuj
– anuj, Commented Jun 26, 2017 at 5:00

airstrike · Accepted Answer · 2017-06-26 05:04:05Z

1

# urls.py
from django.conf.urls import url

from . import views

urlpatterns = [
    url(r'^link/(?P<id>[0-9]+)/$'', views.someview),
]

Then in your views.py

# views.py

def someview(request,id):
    do something with input (id)

This is a very common pattern that is explained thoroughly by the official docs, so I recommend reviewing that.

answered Jun 26, 2017 at 5:04

airstrike

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6 Comments

anuj Over a year ago

( localhost/link/https://google.com ) this is what i am expecting . what will be the url regex to configure it . i dont want id (which is a numerical value )

airstrike Over a year ago

that's not a valid URL, you can't pass that as an argument. are the links like http://google.com stored in your database?

anuj Over a year ago

No its not stored , thats the problem

airstrike Over a year ago

you can encode your url using django.utils.http.urlencode and decode it at the view level, but shit may break in many different ways and can be a security concern. I can't imagine what your use case is that warrants accepting random URLs from users and passing them to a view... I think you might be too concerned about the solution for what you think is your problem that you're missing the forest for the trees, also called the XY problem

anuj Over a year ago

yes , after adding this urls to database , i can get the (id). But it will involve my huge db space . Is there any way so that i can get the solution without involving the database

|

anuj · Accepted Answer · 2017-06-26 09:34:42Z

1

So Finally i have got the ans :

You have to save the the Urls in database as and yo have to point out the id

    a.append(link.objects.get(links=id1))

Then

      {%for i in a %}

    <a href=/link/{{a}}>Click</a><br/>   

     {%endfor%}

answered Jun 26, 2017 at 9:34

anuj

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