Python 3 urllib HTTP错误403：禁止使用参数的localhost

Question

I have been trying to make a POST request in this way:

def simular(request):
    data = urllib.parse.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
    data = data.encode('utf-8')
    request = urllib.request.Request("http://localhost:8000/aguaman/actualizar/")
    # adding charset parameter to the Content-Type header.
    request.add_header("Content-Type","application/x-www-form-urlencoded;charset=utf-8")
    f = urllib.request.urlopen(request, data)
    return HttpResponse("se está simulando la cosa para que funcione: " + str(data))

But I am getting the error 403 forbidden

Then I try to do the same request but I am not using the params so it would be somethink like this:

def simular(request):
    data = urllib.parse.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
    data = data.encode('utf-8')
    request = urllib.request.Request("http://localhost:8000/aguaman/actualizar/")
    # adding charset parameter to the Content-Type header.
    request.add_header("Content-Type","application/x-www-form-urlencoded;charset=utf-8")
    f = urllib.request.urlopen(request)
    return HttpResponse("se está simulando la cosa para que funcione: " + str(data))

And it works, So I would like to know if something is missing, or maybe I cannot use urllib in the way I am doing for local files.

after reading in several places, I found that changing header can help, I already did that and tried to use:

hdr = {
    'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
    'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
    'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
    'Accept-Encoding': 'none',
    'Accept-Language': 'en-US,en;q=0.8',
    'Connection': 'keep-alive'}
    req = urllib.request.Request(url, data, hdr)

But is useless, same result. does anyone can explain the reason? thanks

Answer 1

After making a lot of different changes I started using @csrf_exempt before the views definition, so at the end I finished with these code

from django.views.decorators.csrf import csrf_exempt

@csrf_exempt
def simular(request):
    data = urllib.parse.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
    data = data.encode('utf-8')
    request = urllib.request.Request("http://localhost:8000/aguaman/actualizar/")
    # adding charset parameter to the Content-Type header.
    request.add_header("Content-Type","application/x-www-form-urlencoded;charset=utf-8")
    f = urllib.request.urlopen(request, data)
    return HttpResponse("se está simulando la cosa para que funcione: " + str(data))`

doing this, I was disabling the django security request for this view, I tried to test and the error continued, so I shutted down the server, and turned it on again, and starts working.

Maybe there is something in the server that checks those kind of variables in the app before start running them, so now I can continue working. thanks !