rti*_*dru 9 python django django-models django-related-manager
我有一个抽象基础模型和2个继承模型,我需要强制related_name采用特定格式.
class Animal(models.Model):
legs = models.IntegerField(related_name='%(class)s')
habitat = models.ForeignKey(Habitats, related_name='%(class)s')
class DogAnimal(BaseModel):
name = models.CharField(max_length=20, related_name='dog_animal')
class CatAnimal(BaseModel):
name = models.CharField(max_length=20, related_name='cat_animal')
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通常,related_name ='%(class)s'将分别导致catanimal和doganimal.
我需要像这样的强调值:dog_animal,cat_animal
以下是"为什么"我需要这样做 - 遗产.这些模型没有使用基类组织 - 因此最初指定的related_name是'dog_animal'和'cat_animal'.改变这将是很多工作.
一个解决方案可能不是为所有孩子指定related_nameforhabitat并定义 a default_related_name:
class Animal(models.Model):
class Meta:
abstract = True
habitat = models.ForeignKey(Habitats, on_delete=models.CASCADE)
class DogAnimal(Animal):
class Meta:
default_related_name = 'dog_animal'
class CatAnimal(Animal):
class Meta:
default_related_name = 'cat_animal'
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它需要一些调整,但我认为你可以通过覆盖该类来做到这一点ForeignKey:
from django.utils.text import camel_case_to_spaces
class MyForeignKey(models.ForeignKey):
def contribute_to_class(self, cls, *args, **kwargs):
super().contribute_to_class(cls, *args, **kwargs)
if not cls._meta.abstract:
related_name = self.remote_field.related_name
related_query_name = self.remote_field.related_query_name
underscore_name = camel_case_to_spaces(cls.__name__).replace(" ", "_")
if related_name:
self.remote_field.related_name = related_name.format(
underscore_name=underscore_name
)
if related_query_name:
self.remote_field.related_query_name = related_query_name.format(
underscore_name=underscore_name
)
class Animal(models.Model):
class Meta:
abstract = True
habitat = MyForeignKey(
Habitats, on_delete=models.CASCADE, related_name="{underscore_name}"
)
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