Django 快捷键嵌套外键

Question

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()

我愿意

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

什么是干净的方法？我想到了书上的经理class，但它似乎不适用于这种情况。

当然，我可以在 class 书中实现方法 get_chapters，但我想避免这种情况。

有什么想法吗？

Answer 1

对 FK 字段使用 related_name 参数，结合对查询集的 prefetch_related，将允许您以最小的性能损失获取与书籍相关的所有信息（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，对于最高效的解决方案，还要从章节中保存 FK 到书中。这可以通过覆盖 save 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())

Django 快捷键嵌套外键

Django Shortcut nested foreign key

python

django

nested

foreign-key-relationship