[Django]-Django: passing a url as a parameter

6👍

Firstly, you don’t need to escape the dot if you want it to mean any symbol (in urls.py).

url(r'^show/(?P<url>.+)$', 'myapp.views.json_process')

Secondly, use encodeURIComponent to properly escape the parameter.

var url = 'http://127.0.0.1:8000/show/' +
    encodeURIComponent('http://www.google.com')

By the way, don’t use mixedCase for function names in Python:

Function names should be lowercase, with words separated by
underscores as necessary to improve readability.

Another note that might help in future: don’t hardcode Django URLs in JavaScript. You can generate them dynamically either in your views:

from django.core.urlresolvers import reverse
url = reverse(
    'myapp.views.json_process',
    kwargs={'url': 'http://www.google.com'}
)

{% url myapp.views.json_process url="http://www.google.com" %}