old_contents: """WSGI config for saleor project.
  
  This module contains the WSGI application used by Django's development server
  and any production WSGI deployments. It should expose a module-level variable
  named ``application``. Django's ``runserver`` and ``runfcgi`` commands discover
  this application via the ``WSGI_APPLICATION`` setting.
  
  Usually you will have the standard Django WSGI application here, but it also
  might make sense to replace the whole Django WSGI application with a custom one
  that later delegates to the Django one. For example, you could introduce WSGI
  middleware here, or combine a Django application with an application of another
  framework.
  """
  import os
  
  from django.core.wsgi import get_wsgi_application
  from django.utils.functional import SimpleLazyObject
  
  from saleor.wsgi.health_check import health_check
  
  
  def get_allowed_host_lazy():
      from django.conf import settings
  
      return settings.ALLOWED_HOSTS[0]
  
  
  os.environ.setdefault("DJANGO_SETTINGS_MODULE", "saleor.settings")
  
  # This application object is used by any WSGI server configured to use this
  # file. This includes Django's development server, if the WSGI_APPLICATION
  # setting points here.
  application = get_wsgi_application()
  # Apply WSGI middleware here.
  # from helloworld.wsgi import HelloWorldApplication
  # application = HelloWorldApplication(application)
  application = health_check(application, "/health/")
  
  # Warm-up the django application instead of letting it lazy-load
  application(
      {
          "REQUEST_METHOD": "GET",
          "SERVER_NAME": SimpleLazyObject(get_allowed_host_lazy),
          "REMOTE_ADDR": "127.0.0.1",
          "SERVER_PORT": 80,
          "PATH_INFO": "/graphql/",
          "wsgi.input": b"",
      },
      lambda x, y: None,
  )
  
new_contents: """WSGI config for saleor project.
  
  This module contains the WSGI application used by Django's development server
  and any production WSGI deployments. It should expose a module-level variable
  named ``application``. Django's ``runserver`` and ``runfcgi`` commands discover
  this application via the ``WSGI_APPLICATION`` setting.
  
  Usually you will have the standard Django WSGI application here, but it also
  might make sense to replace the whole Django WSGI application with a custom one
  that later delegates to the Django one. For example, you could introduce WSGI
  middleware here, or combine a Django application with an application of another
  framework.
  """
  import os
  
  from django.core.wsgi import get_wsgi_application
  from django.utils.functional import SimpleLazyObject
  
  from saleor.wsgi.health_check import health_check
  
  
  def get_allowed_host_lazy():
      from django.conf import settings
  
      return settings.ALLOWED_HOSTS[0]
  
  
  os.environ.setdefault("DJANGO_SETTINGS_MODULE", "saleor.settings")
  
  # This application object is used by any WSGI server configured to use this
  # file. This includes Django's development server, if the WSGI_APPLICATION
  # setting points here.
  application = get_wsgi_application()
  # Apply WSGI middleware here.
  # from helloworld.wsgi import HelloWorldApplication
  # application = HelloWorldApplication(application)
  application = health_check(application, "/health/")
  
  # Warm-up the django application instead of letting it lazy-load
  application(
      {
          "REQUEST_METHOD": "GET",
          "SERVER_NAME": SimpleLazyObject(get_allowed_host_lazy),
          "REMOTE_ADDR": "127.0.0.1",
          "SERVER_PORT": 80,
          "PATH_INFO": "/graphql/",
          "wsgi.input": b"",
          "wsgi.multiprocess": True,
      },
      lambda x, y: None,
  )
  
subject: Fix passing wsgi.multiprocess key to the request