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python - 如何在 Django 模型中存储函数

转载 作者:太空宇宙 更新时间:2023-11-04 01:34:31 24 4
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编辑:我完全重写了问题,因为原来的问题没有清楚地解释我的问题

我想运行一个特定于每个特定模型实例的函数。

理想情况下我想要这样的东西:

class MyModel(models.Model):
data = models.CharField(max_length=100)
perform_unique_action = models.FunctionField() #stores a function specific to this instance

x = MyModel(data='originalx', perform_unique_action=func_for_x)
x.perform_unique_action() #will do whatever is specified for instance x

y = MyModel(data='originaly', perform_unique_action=func_for_y)
y.perform_unique_action() #will do whatever is specified for instance y

但是没有数据类型 FunctionField。通常这可以通过继承和创建 MyModel 的子类来解决,可能像这样:

class MyModel(models.Model):
data = models.CharField(max_length=100)
perform_unique_action = default_function

class MyModelX(MyModel):
perform_unique_action = function_X

class MyModelY(MyModel):
perform_unique_action = function_Y

x = MyModelX(data='originalx')
x.perform_unique_action() #will do whatever is specified for instance x

y = MyModelY(data='originaly')
y.perform_unique_action() #will do whatever is specified for instance y

不幸的是,我不认为我可以使用继承,因为我试图以这种方式访问​​函数:

class MyModel(models.Model):
data = models.CharField(max_length=100)
perform_unique_action = default_function

class SecondModel(models.Model):
other_data = models.IntegerField()
mymodel = models.ForeignKey(MyModel)

secondmodel = SecondModel.objects.get(other_data=3)
secondmodel.mymodel.perform_unique_action()

问题似乎是,如果我重写子类中的 perform_unique_action,我不知道外键在 SecondModel 中的类型。

我能否从 SecondModel 访问 MyModel 作为外键,并且仍然为 MyModel 的每个实例提供唯一的函数?

最佳答案

这对我有用。我还没有测试过它,但你应该能够创建另一个类并覆盖它们的方法,它会起作用。检查 class Meta 行,它会将其视为抽象类。这是我现在正在处理的实际类(class)的示例。

编辑:添加了 VoteComment 类并对其进行了测试。它按预期工作!

class Vote(models.Model):
VOTE_ENUM = (
(VoteEnum.DOWN_VOTE, VoteEnum.toString(VoteEnum.DOWN_VOTE)),
(VoteEnum.NONE, VoteEnum.toString(VoteEnum.NONE)),
(VoteEnum.UP_VOTE, VoteEnum.toString(VoteEnum.UP_VOTE)),

)
question = models.ForeignKey(Question, null=False, editable=False, blank=False)
voter = models.ForeignKey(User, blank=False, null=False, editable=False)
vote_type = models.SmallIntegerField(default=0, null=False, blank=False, choices=VOTE_ENUM)

class Meta:
abstract = True

def is_upvote(self):
return self.vote_type > 0
def is_downvote(self):
return self.vote_type < 0

class VoteAnswer(Vote):
answer = models.ForeignKey(Answer, null=False, editable=False, blank=False)

class Meta:
unique_together = (("voter", "answer"),) # to prevent user from voting on the same question/answer/comment again

def __unicode__(self):
vote_type = "UP" if vote_type > 0 else ("DOWN" if vote_type < 0 else "NONE")
return u"{0}: [{1}] {2}".format(user.username, vote_type, answer.text[:32])

def is_upvote(self):
return "FOO! "+str(super(VoteAnswer, self).is_upvote())

class VoteComment(Vote):
comment = models.ForeignKey(Comment, null=False, editable=False, blank=False)

class Meta:
unique_together = (("voter", "comment"),) # to prevent user from voting on the same question/answer/comment again

def __unicode__(self):
vote_type = "UP" if vote_type > 0 else ("DOWN" if vote_type < 0 else "NONE")
return u"{0}: [{1}] {2}".format(user.username, vote_type, comment.text[:32])

def is_upvote(self):
return "BAR!"

关于python - 如何在 Django 模型中存储函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11120652/

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