访问Django模型的超类 [英] Access the superclass of a Django Model

问题描述

给出Django模型

class Sub(models.Model):
    name = models.CharField(max_length=100)
    size_in_inches = models.IntegerField(default=6)

class TunaSub(Sub):
    fish_ingredient = models.CharField(max_length=10, default="Tuna")

class MeatballSub(Sub):
    meat_ingredient = models.CharField(max_length=20, default="Meatball with Cheese")

我想访问超类的属性，例如 __ str __ 方法（在Python 3中）。 X）。我该怎么办？这是正确的解决方案吗？

I would like to access the attribute of the superclass for, say a __str__ method (in Python 3.x). How can I do so? Is this the correct solution?

class TunaSub(Sub):
    fish_ingredient = models.CharField(max_length=10, default="Tuna")
    def __str__(self):
        return self.super().name

class MeatballSub(Sub):
    meat_ingredient = models.CharField(max_length=20, default="Meatball with Cheese")
    def __str__(self):
        return self.super().name

推荐答案

由于您扩展了 Sub ，因此名称也是 TunaSub 和 MeatballSub 的字段。因此，您可以简单地使用

Since you extend Sub, name is also a field of both TunaSub and MeatballSub. So you can simply use

def __str__(self):
    return self.name

作为补充说明，由于要扩展具体模型，因此实际上是在数据库中创建三个单独的表（名为 sub ， tuna_sub 和 meatball_sub ）通过以下方式连接：一对一的关系。如果只想重用 sub 中的字段定义，而不是真正为其创建表，请使用抽象基础模型类。

As a side note, since you are extending a concrete model, you are in fact creating three separate tables in the database (named sub, tuna_sub, and meatball_sub) which are connected via one-to-one relations. If you only want to reuse the field definitions in sub and not actually create a table for it, use an abstract base model class.

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