Как использовать одно приложение в двух разных проектах в Django

'''I made testapp application in baseproject here I create urls.py in testapp and include in base project , Now  I just copy paste the testapp in derivedproject with all urls.py file and views.py , when I add the testapp urls in derived project urls.py using include function it is showing error... 
I want to inform you that I made two projects in same directory and manage.py is inside the projects

Now I am pasting the whole error.....'''
PS C:\Users\hp\Desktop\For Django\day2\derivedproject> python manage.py runserver
Watching for file changes with StatReloader
Exception in thread django-main-thread:
Traceback (most recent call last):
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\threading.py", line 954, in _bootstrap_inner    self.run()
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\threading.py", line 892, in run
    self._target(*self._args, **self._kwargs)
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\utils\autoreload.py", line 64, in wrapper
    fn(*args, **kwargs)
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\core\management\commands\runserver.py", line 110, in inner_run
    autoreload.raise_last_exception()
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\utils\autoreload.py", line 87, in raise_last_exception
    raise _exception[1]
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\core\management\__init__.py", line 375, in execute
    autoreload.check_errors(django.setup)()
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\utils\autoreload.py", line 64, in wrapper
    fn(*args, **kwargs)
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\__init__.py", line 24, 
in setup
    apps.populate(settings.INSTALLED_APPS)
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\apps\registry.py", line 91, in populate
    app_config = AppConfig.create(entry)
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\site-packages\django\apps\config.py", line 224, in create
    import_module(entry)
  File "C:\Users\hp\AppData\Local\Programs\Python\Python39\lib\importlib\__init__.py", line 127, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
  File "<frozen importlib._bootstrap>", line 1030, in _gcd_import
  File "<frozen importlib._bootstrap>", line 1007, in _find_and_load
  File "<frozen importlib._bootstrap>", line 984, in _find_and_load_unlocked
ModuleNotFoundError: No module named 'testapp'
  • введите код здесь

    #derivedproject - файл settings.py установленные приложения INSTALLED_APPS = [ 'django.contrib.admin', 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.messages', 'django.contrib.staticfiles', 'testapp', ]

    #derivedprojects- urls.py

    .

    from django.contrib import admin from django.urls import path,include

    urlpatterns = [ path('admin/', admin.site.urls), path('testapp/',include('testapp.urls')) ]

    #testapp-urls.py

    from django.urls import path from . import views

    здесь находятся урлы уровня приложения

    urlpatterns = [ path('first/', views.first_view), path('second/', views.second_view), path('third/', views.third_view), путь('четвертый/', views.fourth_view), path('fifth/', views.fifth_view), ]

    #derivedproject testapp views.py

     from django.shortcuts import render
     from django.http import HttpResponse
     def first_view(request):
         return HttpResponse('<h1>first view response</h1>')
    
     def second_view(request):
         return HttpResponse('<h1>Second view response</h1>')
    
     def third_view(request):
         return HttpResponse('<h1>Third view response</h1>')
    
     def fourth_view(request):
         return HttpResponse('<h1>fourth view response</h1>')
    
     def fifth_view(request):
         return HttpResponse('<h1>fifth view response</h1>')
    
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