Use only some parts of Django?

Question

I like Django, but for a particular application I would like to use only parts of it, but I\'m not familiar enough with how Django works on the inside, so maybe someone can

Answer 1

I found KeyboardInterrupt's answer but it was answered in 2009 and I failed to run it in Django 1.8.For recent Django 1.8, You can have a look at this, in which some parts come from KeyboardInterrupt's answer.

The folder structure is:

.
├── myApp
│   ├── __init__.py
│   └── models.py
└── my_manage.py

myApp is a module, contains an empty __init__.py and models.py.

There is an example model class in models.py: from django.db import models

class MyModel(models.Model):
     field = models.CharField(max_length=255)

my_manage.py contains django database, installed_app settings and acts as django offical manage.py, so you can:

python my_manage.py sql myApp
python my_manage.py migrate
......

The codes in my_manage.py are: from django.conf import settings

db_conf = {
    'default': {
        'ENGINE': 'django.db.backends.mysql',
        'NAME': 'your_database_name',
        'USER': 'your_user_name',
        'PASSWORD': 'your_password',
        'HOST': 'your_mysql_server_host',
        'PORT': 'your_mysql_server_port',
    }
}

settings.configure(
    DATABASES = db_conf,
    INSTALLED_APPS     = ( "myApp", )
)

# Calling django.setup() is required for “standalone” Django u usage
# https://docs.djangoproject.com/en/1.8/topics/settings/#calling-django-setup-is-required-for-standalone-django-usage
import django
django.setup()

if __name__ == '__main__':
    import sys
    from django.core.management import execute_from_command_line

    execute_from_command_line(sys.argv)