Python Django如何访问抽象基模型局部变量
我已将此抽象基础模型定义如下:Python Django如何访问抽象基模型局部变量,python,django,abstract-class,Python,Django,Abstract Class,我已将此抽象基础模型定义如下: class ActivityAbstractBaseModel(models.Model): POOR = 'PR' FAIR = 'FA' MEDIOCRE = 'ME' GOOD_ENOUGH = 'GE' GOOD = 'GO' VERY_GOOD = 'VG' EXCELLENT = 'EX' STATE = [ (POOR, 'Poor'), (FAIR,
class ActivityAbstractBaseModel(models.Model):
POOR = 'PR'
FAIR = 'FA'
MEDIOCRE = 'ME'
GOOD_ENOUGH = 'GE'
GOOD = 'GO'
VERY_GOOD = 'VG'
EXCELLENT = 'EX'
STATE = [
(POOR, 'Poor'),
(FAIR, 'Fair'),
(MEDIOCRE,'Mediocre' ),
(GOOD_ENOUGH, 'Good Enough' ),
(GOOD, 'Good'),
(VERY_GOOD, 'Very Good'),
(EXCELLENT, 'Excellent'),
]
speaking = models.CharField(max_length=50, choices=STATE, default=GOOD)
然后,我继承了这个抽象模型,如下所示,并添加了新字段writing
class Fluency(ActivityAbstractBaseModel):
writing = models.CharField(max_length=50, choices=STATE, default=GOOD)
现在,这个新字段
writing
正在尝试访问在抽象类中创建的变量GOOD
和STATE
,但是我遇到了namererror
异常。有办法获取这些变量吗?虽然您的类将继承这些类变量,但代码不能直接引用您定义类的位置(因为它们不在该范围内)。相反,您可以在类声明中将它们称为ActivityAbstractBaseModel.GOOD
,等等
class Fluency(ActivityAbstractBaseModel):
writing = models.CharField(max_length=50, choices=ActivityAbstractBaseModel.STATE, default=ActivityAbstractBaseModel.GOOD)
print(Fluency.GOOD) # This works properly
这很奇怪,因为我可以从抽象基础模型访问
speaking
,并使用speaking=None
修改它,甚至删除它。是Django公开了字段而不是其他变量吗?@EyongKevinEnowanyo与Django无关。用普通的类试试看,这样不行。您可以设置speaking
,但这只是您重新定义它,而不是引用它(您可能正在谈论在方法中引用它,该方法也适用于其他变量,例如self。GOOD
在方法中可以正常工作)