在Django模型上定义排序键 [英] Define sorting key on Django model

问题描述

我有两个Django模型，其中一个引用另一个.我希望能够根据第一个模型对第二个模型进行排序.我认为下面的代码示例可以最好地说明我想要实现的目标.

I have two Django models, in which the one refers to the other. I want to be able to sort the second model based on the first. I think the code example below illustrates best what I want to achieve.

class Record(models.Model):
    uuid = models.CharField(max_length=32)
    code = models.CharField(max_length=32)

    class Meta:
        ordering = ['code']

class Article(models.Model):
    code = models.CharField(max_length=32)

    def get_sorted_submodels(self):
        return sorted(self.submodels.all(), key=Submodel.key_sorting)

class Submodel(models.Model):
    code = models.CharField(max_length=32)
    article = models.ForeignKey(Article, related_name='submodels')
    record_uuid = models.CharField(max_length=32)

    @property
    def record(self):
        return Record.objects.get(uuid=self.record_uuid)

    @staticmethod
    def key_sorting(obj):
        return (obj.record, obj.code)

如果现在调用方法get_sorted_submodels，则会出现以下错误:

If I now call the method get_sorted_submodels, I get the following error:

TypeError: unorderable types: Record() < Record()

我已经在模型Record上实现了字段排序.

I already implemented the field ordering on model Record.

1. 如何使该模型可排序，以便可以使用此排序 机制?
2. 如果这不可能，是否还有另一种好方法 启用先按记录"排序，然后再按其自身的代码排序?

1. How can I make this model orderable such that I can use this sorting mechanism?
2. If that is not possible, is there another good way to enable sorting on Record first, and then on its own code?

PS:我明确不想在Submodel类上使用class Meta和ordering，因为这基本上是仅在此上下文中使用的第二顺序.

PS: I explicitly do not want to use class Meta and ordering on Submodel class, since this is basically a second ordering only used in this context.

推荐答案

ordering元字段仅控制在数据库查询中对记录进行排序的方式. sorted是Python函数，与此完全无关.

The ordering Meta field only controls how records are sorted in database queries. sorted is a Python function and completely unrelated to this.

要使Record实例在Python中可排序，可以给他们提供__lt__方法:

To make Record instance sortable in Python, you can give them a __lt__ method:

def __lt__(self, other):
    return self.code < other.code

现在Python可以对它们进行排序了，您的错误将消失.但是最好让数据库来做，所以根本不要使用sorted:

Now Python can sort them, and your error will be gone. But it's better to let the database do it, so don't use sorted at all:

def get_sorted_submodels(self):
    return self.submodels.order_by('record__code')

要在编辑后执行此操作，我将更改方法(从django.utils.decorators导入cached_property):

to do it after your edit, I'd change the methods like so (import cached_property from django.utils.decorators):

@cached_property
def record(self):
    return Record.objects.get(uuid=self.record_uuid)

@staticmethod
def key_sorting(obj):
    return (obj.record.code, obj.code)

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