I am getting this error :-

                                     You're seeing this error because you 
have DEBUG = True in your Django settings file. Change that to False, and 
Django will display a standard 404 page. 


My views.py file:-

from django.http import HttpResponse
import datetime
from django.template import Template, Context
from django.template.loader import get_template
from django.http import Http404  

tasks = [
    {
        'id': 1,
        'title': u'Buy groceries',
        'description': u'Milk, Cheese, Pizza, Fruit, Tylenol', 
        'done': False
    },
    {
        'id': 2,
        'title': u'Learn Python',
        'description': u'Need to find a good Python tutorial on the web', 
        'done': False
    }
 ]

#task_id=='1'
def api(self,task_id):
    #task=[task if task['id']=='task_id' for task in tasks ]
    task = [task for task in tasks if task['id'] == task_id]
    if len(task) == 0:
        raise Http404
    return HttpResponse(tasks)

 My urls.py file:-


"""Dj1 URL Configuration

The `urlpatterns` list routes URLs to views. For more information please 
see:
    https://docs.djangoproject.com/en/1.8/topics/http/urls/
Examples:
Function views
    1. Add an import:  from my_app import views
    2. Add a URL to urlpatterns:  url(r'^$', views.home, name='home')
Class-based views
    1. Add an import:  from other_app.views import Home
    2. Add a URL to urlpatterns:  url(r'^$', Home.as_view(), name='home')
Including another URLconf
    1. Add an import:  from blog import urls as blog_urls
    2. Add a URL to urlpatterns:  url(r'^blog/', include(blog_urls))
"""
from django.conf.urls import include, url
from django.contrib import admin
#from App1.views import place_order_form
from App1.views import api

urlpatterns = [
    #url(r'^admin/', include(admin.site.urls)),
    url(r'^api/(\d{1,2})/$',api),
    #url(r'^place_order/$',place_order_form),
]

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