如何定义Django视图

Question

I am trying to develop a chatbot using Django and depending on the user's input, various python scripts need to be run.

Having the project structure (presented below), is there a way to call the news.py in chatbot.js?

I have tried with an ajax request:

$.ajax({
  type: "POST",
  url: "news/",
  data: { }
}).done(function( o ) {
   print('success')
});

and defined news/ in my urls.py

url('news/', getNews)

where getNews is defined in my views

from artemis.static.marketdata.news import scrape

class getNews():
    scrape()

but I get a 500 error saying that TypeError: object() takes no parameters

Traceback:

Internal Server Error: /news/
 Traceback (most recent call last):
      File "C:\Users\user\AppData\Local\Programs\Python\Python36\lib\site-packages\django\core\handlers\exception.py", line 41, in inner
        response = get_response(request)
      File "C:\Users\user\AppData\Local\Programs\Python\Python36\lib\site-packages\django\core\handlers\base.py", line 187, in _get_response
        response = self.process_exception_by_middleware(e, request)
      File "C:\Users\user\AppData\Local\Programs\Python\Python36\lib\site-packages\django\core\handlers\base.py", line 185, in _get_response
        response = wrapped_callback(request, *callback_args, **callback_kwargs)
    TypeError: object() takes no parameters

What would be the preferable approach in this case? Any tip would be greatly appreciated!

Answer 1

Django has "function" view and "class" view.

You are defining class getNews() . So you have to choose use "function" view or "class" view.

To use function view, change "class" to "def":

from django.http import HttpResponse


def getNews(request):
    result = scrape()
    return HttpResponse(result)

To use class view, define the "getNews()" in correct way.

in view:

from django.views import View
from django.http import HttpResponse

class NewsView(View):
   ...

   def get(self, request, *args, **kwargs):
      result = scrape()
      return HttpResponse(result)

in urls.py:

from views import NewsView
...
url(r'^news/$', NewsView.as_view()),

Answer 2

I think your class declaration is wrong.

class getNews(object):
    def __init__(self, *args, **kwargs):
        # continue here

    def scrape(self):
        # continue here

New-style classes inherit from object , not from NoneType .

---

Note: this answer addresses a code error in the source, nothing to do with django, just python syntax.

Samuel Chen's answer addresses how you create a view in django. A django view is a new-style class, derived from django.views.View . It contains methods corresponding to the HTTP methods (ie get , post , put etc.).