Django running wrong view method - django

Question

I have to view files stored in mysite folder. one is named as views.py and other is named as request_view.py. In urls.py, I have used 'answer' method for views.py and 'display_meta' method for request_view.py.
(django version: 1.5 and python version: 2.7.3)
this is the url pattern:
url(r'^twitter/$', answer), url(r'request/$', display_meta)
when I call http:/127.0.0.1:8000/request/, then also first view(i.e. /twitter/) is called!
any help?
one more thing. In my view.py, I have some unbounded code (i.e. the code which is neither present in a method or class). can this be the cause of the problem?
l = StdOutListener()
auth = OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_token, access_token_secret)
stream = Stream(auth, l)
keyword = input('enter the keyword you want to search for?')
stream.filter(track = [keyword])
apart from this code, evry code is either in the class or method.
One thing that I noticed is that first the code of the view.py runs, then display_meta runs.
Thanks in advance.
SOLVED
The problem was with the import function that I was using. since my code was unbounded in one of the views, the import function always import that regardless of the url that I choose.
Suggestion
Always use the nomenclature mentioned in the this example. In many books it has been suggested that we should import the views, but it might cause an error if you have unbounded code like I had.

Answer 1

I don't know exactly why /twitter/ view is called, but I can see two things to change:
You should use a string as the second parameter for url(), as you can see in this example [1]. You can use 'myapp.views.my_method' nomenclature.
You forgot to start the request URL with ^ that indicates the start of the URL.
About the unbounded code, I don't know if that could be causing the problem. But I can't see why are you putting that code unbounded. I am not sure when that code would be executed, I guess the first time you call a view in that file and Django loads the file (I'm guessing, I don't know exactly), but I don't think that would be a good way to do that. Think when do you want to execute that code, put it in a method, and call it.
[1] https://docs.djangoproject.com/en/1.5/topics/http/urls/#example

Answer 2

HI hemant i am wondering why you have written request_view.py.
Please see the django docs.
what you can do is .
Create two function in your views.py like
def answer(request):
do some stuffs.
render_to_response(template.html)
and on the same page write another
def display_meta(request):
# do your studd
render_to_response(some.html)
YOU NEED NOT TO CREATE TWO SEPERATE VIEWS.PY
I dont know what this code does
l = StdOutListener()
auth = OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_token, access_token_secret)
stream = Stream(auth, l)
keyword = input('enter the keyword you want to search for?')
stream.filter(track = [keyword])
But if you want to use StdOutListener inside your function you can call in your view
like
def display_meta(request):
stobject = StdOutListener() # use parameters if you have here
# do your studd
render_to_response(some.html)

Answer 3

Organizing views into a python package could solve this problem. So if you have a structure like this...
# views.py
def SomeView(request):
return HttpResponse('SomeView')
# another_view.py
def AnotherView(request):
return HttpResponse('AnotherView')
Your can reorganize these separate view files into a views package. That is...
# views
# __init__.py
from views import SomeView
from another_view import AnotherView
# views.py
def SomeView(request):
return HttpResponse('SomeView')
# another_view.py
def AnotherView(request):
return HttpResponse('AnotherView')
And now, everything can be called in a django-standard way:
url(r'^url-to-some-view/$', 'views.SomeView'),
url(r'^url-to-another-view/$', 'views.AnotherView'),
UPDATED:
To make a 'python package'...
Create a views directory at the same level as the view.py file [mkdir views]
Create a __init.py__ file inside the views directory # this is what makes a directory a 'python package'
Move views.py into the views directory.
Move your request_view.py into the views directory.
Edit the __init__.py file with the necessary import statements. In this case:
from views import answer
from request_view import display_meta
What this does is replace a file with a directory. By importing everything into the __init__.py file, this directory looks like a large file to your code, rather than another module.