convert requests.models.Response to Django HttpResponse

Question

In my Django project, I need to get/post some data to a third-party url in my view, and redirect to the web page it provides. For example, I can simply do something like

class TestView(TemplateView):
    def get(self, request, *args, **kwargs):
        data = {
            'order_id': 88888,
            'subject': 'haha',
            'rn_check': 'F',
            'app_pay': 'T',
        }
        url = 'http://some-third-party-api-url?order_id=88888&subject=haha&...'
        return HttpResponseRedirect(url)

However I want to use this third-party api as a wrapped SDK , like

class TestView(TemplateView):
    def get(self, request, *args, **kwargs):
        from sucre.alipay_sdk.base import Alipay
        from sucre.alipay_sdk import alipay_config
        from django.http import HttpResponse
        alipay = Alipay(alipay_config)
        data = {
            'order_id': 88888,
            'subject': 'haha',
            'rn_check': 'F',
            'app_pay': 'T',
        }
        '''alipay api is wrapped in a sdk'''
        '''and return a requests.models.Response instance'''
        result = alipay.api('pay', data)
        return HttpResponse(result)

and the api code:

def api(self, service, data):
    ''' some logics here '''
    import requests
    response = requests.get(url, data=data)
    return response

But seems HttpResponse(result) is not the correct way to convert a requests.models.Response instance to HttpResponse… The layout is bad, and some more encoding issues, etc…Is there a correct way to convert requests response to Django HttpResponse?

Updates:

HttpResponse(result) worked, but some css of the page was lost. This might be related with using requests.

Asked By: Ella SUN

||

Source

Answer 1

This may help you :

requests.models.Response class which,has json() method (according to the documentation) that deserializes the JSON response into a Python object using json.loads(). Try to print following and you can access whatever you are looking for.

print yourResponse.json()

Answered By: Piyush S. Wanare

Answer 2

This should works:

from django.http import HttpResponse
import requests

requests_response = requests.get('/some-url/')

django_response = HttpResponse(
    content=requests_response.content,
    status=requests_response.status_code,
    content_type=requests_response.headers['Content-Type']
)

return django_response

Answered By: paivatulio

Answer 3

To add to paivatulio’s answer, you can forward the headers like so:

for k, v in requests_response.headers.items():
  django_response[k] = v

Answered By: Brian Loughnane

Answer 4

To add to Brian Loughnane’s answer: when I tried the solution:

for k, v in requests_response.headers.items():
    django_response[k] = v

I got an error from django: AssertionError: Hop-by-hop headers not allowed

I don’t know if it’s the best solution but I "fixed" it by removing the offending headers.

from wsgiref.util import is_hop_by_hop

for k, v in requests_response.headers.items():
    if not is_hop_by_hop(k):
        django_response[k] = v

Answered By: Greg Butler

Answer 5

I don’t know if this should be a comment but I don’t have enough reputation to post comments, but this might help:

If you came searching for Django Rest Framework way of doing this, we can modify @paivatulio’s answer like this:

import requests
from rest_framework.response import Response

requests_response = requests.get('/some-url/')

def covert_to_drf_response(response):
    drf_response = Response(
        data=requests_response.json(), 
      status=requests_response.status_code), 
    )

    return drf_response

drf_response = convert_to_drf_response(requests_response)

This might also be helpful if you prefer to use DRF’s Response over HttpResponse for the reasons stated here

You can also set your content_type as @paivatulio did.

Answered By: Basit Balogun

convert requests.models.Response to Django HttpResponse

Question:

Answers: