how to use django.core.management.sql.sql_create?

I want to use this function: django.core.management.sql.sql_create in my view, to get the "CREATE" statements

the function gets 3 arguments:

app, style, connection

what is "app"?? is it a specific object or just the app name?!

I know style is something to do with colors... I reckon django.core.management.color.colorstyle() should work..

what about connection, how do I get this one?

thanks

=========================== edited from here down

ok, after some time, I figured the things, this is what I ended up with:

def sqldumper(model):
    """gets a db model, and returns the SQL statements to build it on another SQL-able db"""
    #This is how django inserts a new record
    #u'INSERT INTO "polls_poll" ("question", "pub_date") VALUES (GE!!, 2011-05-03 15:45:23.254000)'

    result = "BEGIN;\n"
    #add CREATE TABLE statements
    result+= '\n'.join(sql_create(models.get_app('polls'), color_style(), connections.all()[0]))+"\n"
    result+= '\n'.join(sql_custom(models.get_app('polls'), color_style(), connections.all()[0]))+"\n"
    result+= '\n'.join(sql_indexes(models.get_app('polls'), color_style(), connections.all()[0]))+"\n"
    result+= '\n'

    #add INSERT INTO staetements
    units = model.objects.all().values()
    for unit in units:
        statement = "INSERT INTO yourapp.model "+str(tuple(unit.keys()))+" VALUES " + str(tuple(unit.values()))+"\n"
        result+=statement

    result+="\nCOMMIT;\n"
    return 开发者_如何学Cresult.encode('utf-8')

it's still a bit weird, because you get the CREATE TABLE for the whole app, but the INSERT INTO only for the model you ask.... but it's fixable from here

---

It's the models.py module found in one of your apps:

INSTALLED_APPS = (
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.sites',
    'myproject.myapp',
    )

When I printed it, I got < module 'myproject.myapp.models' from '...' >. I used the ellipses instead of typing the entire models.py file's path.