使用 prefetch_related 对多个外键进行 Django 查询集

Question

这是我的模型：

class Owner():
    last_name = models.CharField(max_length=30)
    ...

class Species():
    species_code = models.CharField(max_length=10)
    ...

class Pet():
    client = models.ForeignKey(Owner, related_name='pet_fk')
    species = models.ForeignKey(Species)
    ....

我想列出所有主人及其宠物。有些主人没有宠物，有些主人有很多。

如果找到宠物，我想在该动物的对象上注释一个额外的“临时”字段css_species_class。如果宠物模型的 species_code 是“CANINE”，则该字段将返回“dog”，如果“EQUINE”等，则返回“horse”。

由于站点是多语言的，因此需要“临时”字段，并且需要 css_species_class 值才能在模板中提取适当的字形图标。我不能直接使用存储的值，所以我需要插入一个特定的值来匹配字形所期望的值。

类似：

Owner: John Smith
Pet: Saag (css_species_class='dog')
Pet: Brinjal (css_species_class='cat')
Pet: Baji (css_species_class='dog')

Owner: Sue Smith
Pet: none

Owner: Clare Smith
Pet: Aloo (css_species_class='horse')

我的模板是这样的：

{% for owner in owners %}
    <tr>
        <td>{{ owner.first_name }} {{ owner.last_name }}</td>
        <td> <!-- loop over pet objects -->
            {% for pet in owner.pet_fk.all %}
                <div>
                    ....
                    <span class="glyphicons glyphicons-{{ pet.css_species_class }}"></span>
                    ....
                </div>
            {% endfor pet %}
        </td>
    </tr>
{% endfor %}

所以，这是我第一次尝试解决方案：

class OwnerListView(ListView):
    template_name = 'visitors/owner_list.html'
    context_object_name = 'owners'
    paginate_by = 50

    def get_queryset(self):
        owners_with_pets = Owner.objects.filter(pet_fk__isnull=False).prefetch_related('pet_fk').distinct()
            # logic goes here to loop over pets
            # and assign 'css_species_class' temp field

        owners_without_pets = Owner.objects.filter(pet_fk__isnull=True).prefetch_related('pet_fk').distinct()

然后将两个查询集“合并”在一起：

        result_list = sorted(
            chain(owners_with_pets, owners_without_pets),
            key=attrgetter('last_name'))
            return result_list

这对少数所有者“有效”，但如果我使用实数（约 4,000）进行测试，我会收到“sql 变量太多”错误。

我最初尝试在单个查询中执行此操作（在决定将其分解为两个查询之前），但对于大量客户来说同样失败了。

有人可以就如何最好地解决这个问题给我一些指导吗？非常感谢。

Answer 1

试试这个代码，没有经过测试，但我认为它可以工作。

class Owner():
    last_name = models.CharField(max_length=30)
    ...

class Species():
    species_code = models.CharField(max_length=10)
    ...

class Pet():
    # related_name is used for backward relation
    # this will end up as owner."related_name" --> owner.pets
    client = models.ForeignKey(Owner, related_name='pets')
    species = models.ForeignKey(Species)
    ...

    @property
    def css_species_class(self):
        # this could be anything you want eg: css_scecies_class
        return self.species.species_code


class OwnerListView(ListView):
    template_name = 'visitors/owner_list.html'
    context_object_name = 'owners'
    paginate_by = 50

    def get_queryset(self):
        # no need to check if onwner instance has pets if you chain them both back
        return Owner.objects.prefetch_related('pets').all().distinct()

{% for owner in owners %}
    <tr>
        <td>{{ owner.first_name }} {{ owner.last_name }}</td>
        <td> <!-- loop over pet objects -->
            {% for pet in owner.pets %}
                <div>
                    ....
                    <!-- now we can access pet.css_species_class directly because we made it a property of pet class -->
                    <span class="glyphicons glyphicons-{{ pet.css_species_class }}"></span>
                    ....
                </div>
            {% endfor pet %}
        </td>
    </tr>
{% endfor %}