Django 快捷方式嵌套外键



>假设我的 models.py 中有以下内容:

class Book:
    pass
class Part:
    book = models.ForeignKey(Book)
class Chapter:
    part = models.ForeignKey(Part)
    number = models.IntegerField()

我想做

book = Book.objects.get(id=someID)
chapters = Book.chapters.get(number=4)

什么是干净的方法?我想到了书课上的经理,但它似乎不适用于这种情况。

当然,我可以在课本上实现get_chapters的方法,但我想避免这种情况。

有什么想法吗?

对 FK 字段使用 related_name 参数,再加上对查询集的prefetch_related,将允许您获取与书籍相关的所有信息,同时对性能的影响最小(每个prefetch_related参数调用单独的查询)。

class Book:
    pass
class Part:
    book = models.ForeignKey(Book, related_name="parts")
class Chapter:
    part = models.ForeignKey(Part, related_name="chapters")
    number = models.IntegerField()
# fetch a book and all related info w/ only 2 db hits
book = Book.objects.first().prefetch_related("parts","parts__chapters")
print(book.parts.all()) # returns all parts for book
for part in book.parts.all():
    print part.chapters.all()

您也可以在模板中执行此操作。

但是,对于性能最高的解决方案,也保存 FK 以从章节预订。这可以通过重写保存方法轻松完成。

class Chapter:
    part = models.ForeignKey(Part, related_name="part_chapters")
    number = models.IntegerField()
    book = models.ForeignKey(Book, related_name="chapters", null=True, blank=True) # allow null/blank values; will be populated in save method
    def save(self, *args, **kwargs):
        self.book = self.part.book
        super(Chapter, self).save(*args, **kwargs)
>>> book = Book.objects.first().prefetch_related("parts","chapters")
>>> print(book.parts.all()) # returns all parts for book
>>> print(book.chapters.all())

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