我以某种模式创建了一组模型,以便某些模型可以包含来自多个父类的详细信息。我的项目需要我有这些主要类别:Django:具有多个行内表的父表使用外键从父项链接到每个行内表
-- {class name} {file name they are contained in}
-- Attraction (attraction.py)
-- Tours (tours.py)
-- Transit (transit.py)
-- Flight (flight.py)
-- Accomodation (accomodation.py)
他们每个人都有多个公共领域,其建模成普通机型;我想有从班外键上方到下方的那些:
-- Contact (contact.py)
-- Address (address.py)
-- LocationMetadata (location.py)
-- Duration (duration.py)
-- Price (price.py)
我从
attraction --> contact, address, location, price
transit --> location, price
accomodation --> location, type, price, contact, address
范例模型Atraction取得外键:
class Attraction(models.Model):
...
ATTRACTION_TYPE = (
('NIGHT-LIFE', 'NL'),
('EARTERY', 'EAT'),
('CULTURAL', 'CUL'),
('LANDMARK', 'LNDMRK'),
('ADVENTURE', 'ADV'),
('ENTERTAINMENT', 'ENT'),
('SCENIC', 'SNC'),
('WILDLIFE', 'WDL')
)
type = forms.MultipleChoiceField(choices=ATTRACTION_TYPE)
...
我想创建管理员访问权限(吸引力,过境,住宿)作为接入点和休息,以便与他们联系。但是这在Django中是不允许的...有没有办法绕过它?
张贴您的模特。没有他们很难提供帮助。 – 2014-11-25 11:38:33