django ignore slash in url and take parameter

Question

hi i have url like this:

path('api/v1/store/download/<str:ix>/', DownloadVideoAPI.as_view(), name='download'),

it accept long string .

I want to keep allthing after download key in above URL as the parameter.

but when I enter a long string that contains some slash Django says page not found for example when if enter "/api/v1/store/download/asdasd2asdsadas/asdasd" will give me 404 not found ...

how can I do that?

this is my view:

class DownloadVideoAPI(APIView):
    def get(self, request, ix):
        pre = ix.split(",")
        hash = pre[0]
        dec = pre[1]
        de_hash = decode_data(hash, dec)

BugHunter · Accepted Answer · 2018-04-28 12:59:04Z

1

Well, It's possible to add the extra parameters in the request. you can use re_path method.

# urls.py
from django.urls import re_path

re_path(r'api/v1/store/download/(?P<ix>\w+)/', DownloadVideoAPI.as_view(), name='download'),

ref: https://docs.djangoproject.com/en/2.0/ref/urls/#django.urls.re_path

edited Apr 28, 2018 at 12:59

answered Apr 28, 2018 at 12:46

BugHunter

1,2042 gold badges10 silver badges15 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

vishes_shell · Accepted Answer · 2018-04-28 12:37:21Z

0

Just use

path('api/v1/store/download/<str:ix>', DownloadVideoAPI.as_view(), name='download'),

without / at the end.

answered Apr 28, 2018 at 12:37

vishes_shell

23.8k7 gold badges74 silver badges82 bronze badges

Comments

Sean Francis N. Ballais · Accepted Answer · 2018-04-28 12:54:01Z

0

/api/v1/store/download/asdasd2asdsadas/asdasd will result in a 404 page since Django cannot map the URL, /api/v1/store/download/asdasd2asdsadas/, to a route in your urls.py. To solve this, aside from using BugHunter's answer, you could URL encode your long string first before passing it to your URL.

So, given the long string, "asdasd2asdsadas/asdasd", URL encode it first to "asdasd2asdsadas%2Fasdasd". Once you have encoded it, your URL should now look like "/api/v1/store/download/asdasd2asdsadas%2Fasdasd".

To URL encode in Python 3, you can use urllib.

import urllib

parameter = 'asdasd2asdsadas/asdasd'
encoded_string = urllib.quote(parameter, safe='')

encoded_string here should have the value, "asdasd2asdsadas%2Fasdasd".

edited Apr 28, 2018 at 12:54

answered Apr 28, 2018 at 12:39

Sean Francis N. Ballais

2,4982 gold badges28 silver badges44 bronze badges

Collectives™ on Stack Overflow

django ignore slash in url and take parameter

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related