Question

这里是一个抽象的简化示例。

说我想找作者并注释类别中最少的书籍（如果该数目大于3）。

书本和作者模型并且与ForeignKey字段无关（请记住，抽象和简化，这是有原因的）：

Author(models.Model):
    name = models.CharField(max_length=250)

Book(models.Model):
    author_name = models.CharField(max_length=250)
    book_category = models.CharField(max_length=250)

这是我可以重现的最简单的查询：

(Author.objects
 .annotate(min_valuable_count=Subquery(
    Book.objects
        .filter(author_name=OuterRef('name'))
        .annotate(cnt=Count('book_category'))
        .filter(cnt__gt=3)
        .order_by('cnt')
        .values('cnt')[:1],
    output_field=models.IntegerField()
 ))
)

我得到一个错误：

psycopg2.ProgrammingError: missing FROM-clause entry for table "U0"
LINE 1: ... "core_author" GROUP BY "core_author"."id", "U0"."id" ...
                                                       ^

这是SQL：

SELECT "core_author"."id", "core_author"."name", (
    SELECT COUNT(U0."book_category") AS "cnt" 
    FROM "core_book" U0 WHERE U0."id" = ("core_author"."chat_id") 
    GROUP BY U0."id" HAVING COUNT(U0."book_category") > 3 
    ORDER BY "cnt" ASC  LIMIT 1) 
AS "min_valuable_count" 
FROM "core_author" 
GROUP BY "core_author"."id", "U0"."id"

更新＃1

我发现删除.filter(cnt__gt=3)会删除无法访问GROUP BY的最后U0：

SELECT "core_author"."id", "core_author"."name", (
    SELECT COUNT(U0."book_category") AS "cnt" 
    FROM "core_book" U0 WHERE U0."id" = ("core_author"."chat_id") 
    GROUP BY U0."id"
    ORDER BY "cnt" ASC  LIMIT 1) 
AS "min_valuable_count" 
FROM "core_author"

是否可以在不删除子查询中的GROUP BY的情况下在外部查询中删除.filter(cnt__gt=3)？

Answer 1

这是bug，将在Django 2.1.6版中修复。现在是工作区：

min_valuable_count_qs = Subquery(
    Book.objects
        .filter(author_name=OuterRef('name'))
        .annotate(cnt=Count('book_category'))
        .filter(cnt__gt=3)
        .order_by('cnt')
        .values('cnt')[:1],
    output_field=models.IntegerField()
)

min_valuable_count_qs.contains_aggregate = False

qs = Author.objects.annotate(min_valuable_count=min_valuable_count_qs))

子查询中带注释的Django过滤器按计数

1 个答案: