CODEWITHSUNDEEP
pythondatabasedjangodjango-modelsdjango-database
LondonAppDev
LondonAppDev

Reputation: 9673

Multiple Django Databases - Map Model to Database in Same Application

I have searched all over for a solution for this but have been unable to find anything.

I have one Django project, one application, two models and two database. I would like one model to speak and sync exclusively to one database. This is what I've tried:

Settings

DATABASES = {
    'default': {
        'ENGINE': 'django.db.backends.mysql', # Add 'postgresql_psycopg2', 'mysql', 'sqlite3' or 'oracle'.
        'NAME': 'database_a',                      # Or path to database file if using sqlite3.
        # The following settings are not used with sqlite3:
        'USER': 'user',
        'PASSWORD': 'xxxxxx',
        'HOST': 'localhost',                      # Empty for localhost through domain sockets or '127.0.0.1' for localhost through TCP.
        'PORT': '',                      # Set to empty string for default.
    },
    'applicationb_db': {
        'ENGINE': 'django.db.backends.mysql',
        'NAME': 'database_b',
        'USER': 'user',
        'PASSWORD': 'xxxxxx',
        'HOST': 'localhost',
        'PORT': '',                 
    },
}
DATABASE_ROUTERS = ['fanmode4.router.ApiRouter']

Models

from django.db import models

class TestModelA(models.Model):
    testid = models.CharField(max_length=200)
    class Meta:
        db_table = 'test_model_a'

class TestModelB(models.Model):
    testid = models.CharField(max_length=200)
    class Meta:
        db_table = 'test_model_b'
        app_label = 'application_b'

Router

class ApiRouter(object):
    def db_for_read(self, model, **hints):
        if model._meta.app_label == 'application_b':
            return 'applicationb_db'
        return None

    def db_for_write(self, model, **hints):
        if model._meta.app_label == 'application_b':
            return 'applicationb_db'
        return None

    def allow_relation(self, obj1, obj2, **hints):
        if obj1._meta.app_label == 'application_b' or \
           obj2._meta.app_label == 'application_b':
           return True
        return None

    def allow_syncdb(self, db, model):
        if db == 'applicationb_db':
            return model._meta.app_label == 'application_b'
        elif model._meta.app_label == 'application_b':
            return False
        return None

the application name is "api". Basically with this setup, if I sync the database it will only sync on the default db. If I sync the database specifying the second database python manage.py syncdb --database=applicationb_db, it will not sync anything to the second database.

I am simply trying to achieve the following:

Upvotes: 1

Views: 1920

Answers (1)

Rohan
Rohan

Reputation: 53386

Instead of using model._meta.app_label you can use model to check which model it is and return appropriate DB.

You can update the router as:

class ApiRouter(object):
    def db_for_read(self, model, **hints):
        if model == TestModelB:
            return 'applicationb_db'
        return None

    def db_for_write(self, model, **hints):
        if model == TestModelB:
            return 'applicationb_db'
        return None

    def allow_relation(self, obj1, obj2, **hints):
        if model == TestModelB:
           return True
        return None

    def allow_syncdb(self, db, model):
        if model == TestModelB:
            return True
        else:
            return False
        return None

Upvotes: 3

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