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()

我想做

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

干净的方法是什么？我在书课上想到了一位经理，但它似乎并不适用于这种情况。

当然我可以在课堂上实现get_chapters方法，但我想避免这种情况。

任何想法？

Answer 1

对FK字段使用related_name参数，再加上查询集的prefetch_related，将允许您以最小的性能命中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，对于性能最佳的解决方案，请将FK保存到章节中。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 Shortcut嵌套外键

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