[Django]-In django, how do I call the subcommand 'syncdb' from the initialization script?

As suggested by “Where to put Django startup code?“, you can use middleware for your startup code. The Django docs are here.

For example (untested):

startup.py:

from django.core.exceptions import MiddlewareNotUsed
from django.conf import settings
from django.core.management import call_command

class StartupMiddleware(object):
    def __init__(self):
        # The following db settings name is django 1.2.  django < 1.2 will use settings.DATABASE_NAME
        if settings.DATABASES['default']['NAME'] == ':memory:':
            call_command('syncdb', interactive=False)

        raise MiddlewareNotUsed('Startup complete')

and in your settings.py:

 MIDDLEWARE_CLASSES = (
     'your_project.middleware.startup.StartupMiddleware',

     # Existing middleware classes here
 )

Update

I added a script called run.sh in the project’s root directory. This worked for me with an SQLite database:

#!/usr/bin/python
from django.core.management import call_command
call_command('syncdb')
call_command('runserver')

Original Answer

I am not sure I understand what you mean by “scripting the syncdb command”. You usually execute python manage.py syncdb from the command line. This is usually done after adding new models. In case you want to you easily accomplish this using a simple shell script. I don’t see any reason to place (or invoke) syncdb from within settings.py.

Could you add more details to your question? Add context and explain what exactly are you trying to get done?

@Daniel Naab’s answer, as well as the doc in the official site, is not for executing management commands as an entrypoint.

When you want to use a management command as the entrypoint in managed cloud environment like AWS Lambda or Google Cloud Functions, you can take a look at manage.py and try something similar.

import os
from django.core.management import execute_from_command_line

def publishing_fn(data, context):
    os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'YOURAPP.settings')
    # The first argument is "manage.py" when it's run from CLI.
    # It can be an empty string in this case
    execute_from_command_line(['', 'COMMAND', 'ARGS...'])

You could create a new script that you call instead of manage.py that calls manage.py:

from subprocess import call
call(["python", "manage.py", "syncdb"])
call(["python", "manage.py", "runserver"])

If you don’t need to add an admin you could change the second line like this:

call(["python", "manage.py", "syncdb", "--noinput"])

I’m assuming that what you’re trying to do is create your db and then start your server with one command every time.