我正在尝试在1个数据库中有选择地创建表,而不是另一个。
在我的示例中,我只想为tlocation
数据库中的应用创建表:tcategory
和transforms
。但是,Django正在transforms
数据库中创建所有表。
这是DB路由器配置:
TRANSFORM_APPS = ('tcategory', 'tlocation')
class TransformRouter(object):
"""
A router to control all database operations on models in the
"utils_transform" application.
"""
def db_for_read(self, model, **hints):
"""
Attempts to read 'transforms' models go to 'transforms' database.
"""
if model._meta.app_label in TRANSFORM_APPS:
return 'transforms'
return None
def db_for_write(self, model, **hints):
"""
Attempts to write 'transforms' models go to 'transforms' database.
"""
if model._meta.app_label in TRANSFORM_APPS:
return 'transforms'
return None
def allow_relation(self, obj1, obj2, **hints):
"""
Allow relations if a model in the 'tlocation' app is involved.
"""
if obj1._meta.app_label in TRANSFORM_APPS or \
obj2._meta.app_label in TRANSFORM_APPS:
return True
return None
def allow_migrate(self, db, app_label, model=None, **hints):
"""
Make sure the 'tlocation' app only appears in the 'transforms'
database.
"""
if app_label in TRANSFORM_APPS:
return db == 'transforms'
return None
class DefaultRouter(object):
"""
Catch-all Router for all other DB transactions that aren't in the
``utils_transform`` app.
"""
def db_for_read(self, model, **hints):
return 'default'
def db_for_write(self, model, **hints):
return 'default'
def allow_relation(self, obj1, obj2, **hints):
if obj1._state.db == obj2._state.db:
return True
return None
def allow_migrate(self, db, app_label, model=None, **hints):
return None
我用来运行迁移的命令是:
./manage.py migrate --database=transforms
当我一次只迁移1个应用时,如下所示,这是有效的。但是如果没有migrate
命令中的应用过滤器,我就无法运行。例如:
./manage.py migrate tlocation --database=transforms
./manage.py migrate tcategory --database=transforms
答案 0 :(得分:2)
您是否尝试过managed = False
您想要创建表格的模型?
class MyModel(models.Model):
...
class Meta:
managed = False