将django RawQuerySet转换为Queryset

Question

我有2个Django模型,ModelA用ArrayField该用于存储的主键值的大列表(可能50K +列表)

class ModelA(models.Model):
    pk_values = ArrayField(models.IntegerField())

class CustomManager(manager.Manager):

    def get_for_index(self, index_id):
        qs = self.get_queryset()
        obj = ModelA.objects.get(pk=index_id)
        return qs.filter(id__in=obj.pk_values)

class ModelB(models.Model):
    # [...] some fields

    objects = CustomManager()

这有效:

qs = ModelB.objects.get_for_index(index_id=1)

但是,如果"pk_values"是一个大型列表,这将是超级慢的.

所以我尝试了原始的SQL查询:

class CustomManager(manager.Manager):
    def get_for_index(self, index_id):
        qs = self.get_queryset()
        sql = "SELECT * FROM myapp_model_b JOIN myapp_model_a ON myapp_model_b.id = ANY(myapp_model_a.pk_values) WHERE myapp_model_a.id = '%s'" % index_id
        return qs.raw(sql)

但这会返回一个django.db.models.query.RawQuerySet实例.

但有了这个,我就不能做事queryset.values()了.

如何将其转换为普通的Django查询集？

有没有更好的方法呢？

文档:

• ArrayField https://docs.djangoproject.com/en/2.0/ref/contrib/postgres/fields/#arrayfield
• 自定义管理器https://docs.djangoproject.com/en/2.0/topics/db/managers/#custom-managers-and-model-inheritance
• 原始查询https://docs.djangoproject.com/en/2.0/topics/db/sql/#performing-raw-sql-queries

Answer 1

您可以使用RawSQL表达式:

ModelB.objects.filter(id__in=RawSQL(
    'SELECT unnest(a.pk_values) FROM app_modela a WHERE a.id = %s',
    [index_id]
))

或者,您可以使用extra()重现您在问题中的确切查询:

ModelB.objects.extra(
    tables=['foo_modela'],
    where=[
        '"app_modelb"."id" = ANY("app_modela"."pk_values")',
        '"app_modela"."id" = %s',
    ],
    params=[index_id],
)