CommandError: Unknown command:

I created a custom command and want to run it automatically on start

My command file

# project/parser_app/management/commands/create_schedules.py

class Command(loaddata.Command):
    def handle(self, *args, **options):
        # some code

Added command to wsgi.py

import os
from django.core.wsgi import get_wsgi_application
from django.core.management import call_command

os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'core_app.settings')

application = get_wsgi_application()
call_command("makemigrations") 
call_command("migrate")
call_command("collectstatic", interactive=False)
call_command("create_schedules parser_app/management/commands/schedule_fixtures.json")

I can run in console command python manage.py "create_schedules parser_app/management/commands/schedule_fixtures.json" and everything is working but when I added command to wsgi.py (to run on start)

import os
from django.core.wsgi import get_wsgi_application
from django.core.management import call_command

os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'core_app.settings')

application = get_wsgi_application()
call_command("makemigrations") 
call_command("migrate")
call_command("collectstatic", interactive=False)
call_command("create_schedules parser_app/management/commands/schedule_fixtures.json")

I have an error

raise CommandError("Unknown command: %r" % command_name) django_1 | django.core.management.base.CommandError: Unknown command: 'create_schedules parser_app/management/commands/schedule_fixtures.json'

UPDATE

I was trying to run command as a function^ but have an error


ile "/project/core_app/wsgi.py", line 14, in <module>
django_1         |     from parser_app.management.commands.start_command import create_schedules
django_1         |   File "/project/parser_app/management/commands/start_command.py", line 1, in <module>
django_1         |     from parser_app.models import Schedule
django_1         |   File "/project/parser_app/models.py", line 7, in <module>
django_1         |     class TimeBase(models.Model):
django_1         |   File "/usr/local/lib/python3.9/site-packages/django/db/models/base.py", line 127, in __new__
django_1         |     app_config = apps.get_containing_app_config(module)
django_1         |   File "/usr/local/lib/python3.9/site-packages/django/apps/registry.py", line 260, in get_containing_app_config
django_1         |     self.check_apps_ready()
django_1         |   File "/usr/local/lib/python3.9/site-packages/django/apps/registry.py", line 138, in check_apps_ready

Incorrect syntax, see below for correction:

call_command("create_schedules", "parser_app/management/commands/schedule_fixtures.json")

Simpler still would be to not have the code as a management command:

# create_schedules.py

def create_schedules(file_path):
   ...

# wsgi.py

import create_schedules as cs

[...]
cs.create_schedules("parser_app/management/commands/schedule_fixtures.json")

Solved with help of ready() function in apps.py

My solution

class ParserAppConfig(AppConfig):
    default_auto_field = 'django.db.models.BigAutoField'
    name = 'parser_app'

    def ready(self):
        from scripts.create_schedules import create_cron_templates

        create_cron_templates()
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