How to make a redirect and keep the query string?

Question

I want to make a redirect and keep what is the query string. Something like self.redirect plus the query parameters that was sent. Is that possible?

Infarct

Where do you want to keep it? In a session? – Vinson

Can't I just pass on the query parameters via HTTP GET? – Infarct

Of course, I don't know which framework you are using, but that should be straightforward. In straight http you would send a 301 or 303 with the Location header set to the redirect url plus the query params you want to keep. – Vinson

Yes, that's what I want to do and I think it is straightforward. I use webapp2 and google app engine, I'm adding those tags to the question. – Infarct

@Wooble I could try the parameters one by one but I'm looking for a way to add the entire query string in a oneliner. I know the names of the parameters I want to get via HTTP GET and I want to pass on their values. I figure there should be a way to pass on the entire map instead of one variable at a time. – Infarct

Answer 1

newurl = '/my/new/route?' + urllib.urlencode(self.request.params)
self.redirect(newurl)

Happen

can not run in tornado 4 – Giacinta

Can use urllib.parse.urlencode(self.request.query_arguments, doseq=True) in then new version. – Counterblow

Answer 2

You can fetch the query string to the current request with self.request.query_string; thus you can redirect to a new URL with self.redirect('/new/url?' + self.request.query_string).

Loidaloin

can not run in tornado 4 – Giacinta

Answer 3

Use the RedirectView.

from django.views.generic.base import RedirectView
path('go-to-django/', RedirectView.as_view(url='https://djangoproject.com', query_string=True), name='go-to-django')

Rosin

Answer 4

This worked for me in Django 2.2. The query string is available as a QueryDict instance request.GET for an HTTP GET and request.POST for an HTTP POST. Convert these to normal dictionaries and then use urlencode.

from django.utils.http import urlencode

query_string = urlencode(request.GET.dict())  # or request.GET.urlencode()

new_url = '/my/new/route' + '?' + query_string

See https://docs.djangoproject.com/en/2.2/ref/request-response/.

Highroad