How can I go to a specific URL with parameters like if I have view
def search(request):
and in urls.py
^search/$
and what I need to do is to redirect like search/?item=4
c = {}
render_to_response("search.html",c)
works fine, but
render_to_response("search.html/?item=" + itemID, c )
it says no template found ( I know there is no template like search.html/?item= ) but how can I pass parameters or use query string to redirect?
Using reverse and passing the name of url we can redirect to url with query string:
#urls.py
url(r'^search/$', views.search, name='search_view')
#views.py
from django.shortcuts import redirect, reverse
# in method
return redirect(reverse('search_view') + '?item=4')
urlencode as in this answer
Location header manually that's true but if you use HttpResponseRedirect or the redirect shortcut, it calls iri_to_uri which calls urlencode for you so you don't need to do it yourself. If fact, if you do it yourself, the URL will get double encoded which mangles it.I know this question is a bit old, but someone will stumble upon this while searching redirect with query string, so here is my solution:
import urllib
from django.shortcuts import redirect
def redirect_params(url, params=None):
response = redirect(url)
if params:
query_string = urllib.urlencode(params)
response['Location'] += '?' + query_string
return response
def your_view(request):
your_params = {
'item': 4
}
return redirect_params('search_view', your_params)
If you already have a request (with a GET containing some parameters), and you want to redirect to a different view with the same args, kwargs and parameters, we can use the QueryDict object to url encode itself:
def view(request, *args, **kwargs):
return redirect(
reverse('your:view:path', args=args, kwargs=kwargs) +
'?' + request.GET.urlencode()
)
QueryDict instance, see e.g. stackoverflow.com/a/15010861A more generic option;
from urllib.parse import urlencode
from django.shortcuts import redirect as django_redirect
def redirect(url, *args, params=None, **kwargs):
query_params = ""
if params:
query_params += '?' + urlencode(params)
return django_redirect(url+query_params, *args, **kwargs)
To redirect to another page while carrying along the the current query strings:
views.py:
from django.urls import reverse
from django.shortcuts import redirect
def my_view(request):
#get the current query string
q = request.META['QUERY_STRING']
return redirect(reverse('search_view') + '?' + q)
