Question

我想写一个方法来返回对象的所有后代（即子代，孙代，曾孙...）作为QuerySet。

我写了两种方法，第一种（get_all_children）可以返回子代和孙代，但是如果有曾孙（或后代）则给出以下错误：

django.db.utils.ProgrammingError: (1064, "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'UNION (SELECT DISTINCT `deneme_app_person`.`id`, `deneme_app_person`.`name`, `de' at line 1")

第二种方法（get_all_children_list）正常工作（据我测试），但可能较慢，最后一行看起来不太好。

models.py

from django.db import models


# Create your models here.
class Person(models.Model):
    name = models.CharField(max_length=100)
    parent = models.ForeignKey(
        'self', null=True, blank=True, related_name='children',
        on_delete=models.CASCADE)

    def get_all_children(self):
        """
        This method will return all children and grandchildren of Person objects
        :return:
        """
        p_list = self.children.all()
        if p_list.count():
            for p in p_list:
                children = p.get_all_children()
                if children.count():
                    p_list = p_list.union(children)
        return p_list.distinct()

    def get_all_children_list(self):
        p_list = self.children.all()
        if p_list.count():
            for p in p_list:
                children = p.get_all_children_list()
                if children.count():
                    p_list = set(p_list).union(children)
                    # p_list = list(set(p_list + children))
        return Person.objects.filter(id__in=[p.id for p in p_list])
        # return list(p_list)

    def __str__(self):
        return f"({self.id}) - {self.name}"

tests.py

from django.test import TestCase
from django.db.models.query import QuerySet

from .models import Person


class PersonTests(TestCase):
    def test_get_all_children(self):
        p = Person.objects.create(name='Jack')
        p1 = Person.objects.create(name='Jack Jr', parent=p)
        p2 = Person.objects.create(name='Jill', parent=p)
        p11 = Person.objects.create(name='Jack Jr2', parent=p1)
        p111 = Person.objects.create(name='Jack Jr3', parent=p11)
        p112 = Person.objects.create(name='Jill Jr', parent=p11)

        # Following test is working
        self.assertEqual([p11, p111, p112], list(p1.get_all_children()))
        # Following test is not working
        self.assertEqual([p1, p2, p11, p111, p112], list(p.get_all_children()))

        for person in [p, p1, p2, p11, p111, p112]:
            self.assertIsInstance(person.get_all_children(), QuerySet)

    def test_get_all_children_list(self):
        p = Person.objects.create(name='Jack')
        p1 = Person.objects.create(name='Jack Jr', parent=p)
        p2 = Person.objects.create(name='Jill', parent=p)
        p11 = Person.objects.create(name='Jack Jr2', parent=p1)
        p111 = Person.objects.create(name='Jack Jr3', parent=p11)
        p112 = Person.objects.create(name='Jill Jr', parent=p11)
        p1111 = Person.objects.create(name='Jack Jr4', parent=p111)

        self.assertEqual([p1111], list(p111.get_all_children_list()))
        self.assertEqual([p111, p112, p1111], list(p11.get_all_children_list()))
        self.assertEqual([p11, p111, p112, p1111], list(p1.get_all_children_list()))
        self.assertEqual([p1, p2, p11, p111, p112, p1111], list(p.get_all_children_list()))

        for person in [p, p1, p2, p11, p111, p112, p1111]:
            self.assertIsInstance(person.get_all_children_list(), QuerySet)

您能指出第一种方法有什么问题吗？如何纠正该方法？

Answer 1

我在本地重现了您的错误，但我还没有弄清楚其背后的原因。如果您不介意不使用union，则可以使用|运算符合并两个查询集，因此您的方法将如下所示：

def get_all_children(self):
    children = self.children.all()
    for child in children:
        children = children | child.get_all_children()

    return children

在Django中获取对象的后代

1 个答案: