sql - 如何获得命名内部联接的非重复计数？

Question

我已经为信息汇总表构建了一个查询，它几乎完成了，但有一个小错误。如果一个类(class)有多个用户，confirmed_class_count 变量会返回太高，这让我相信这个数字不明显

这是我当前的代码:

SELECT "staffs".*,
  count(distinct subclasses) as class_count,
  sum(case when users.confirmed_at is not null then 1 else 0 end) confirmed_class_count
FROM
  staffs
INNER JOIN classes as subclasses on staffs.staff_id = ANY(subclasses.staff)
INNER JOIN "classes_users" ON "classes_users"."class_id" = "subclasses"."id"
INNER JOIN "users" ON "users"."id" = "classes_users"."user_id"
INNER JOIN class_types ON class_types.code = subclasses.class_type_code
WHERE 
  (subclasses.closed_date is NULL OR subclasses.closed_date > '2019-09-06')
GROUP BY
  staffs.id ORDER BY "staffs"."full_name" ASC

我想用 (select distinct count(*) from subcases where users.confirmed_at is not null) as confirmed_case_count 替换 sum 但我得到 关系“子类”不存在。

我怎样才能得到我想要的东西？

Answer 1

您可以将非重复计数与条件聚合结合使用。替换

sum(class when users.confirmed_at is not null then 1 else 0 end) confirmed_class_count
    ^ looks like a typo, this should be case not class

与

count(distinct case when users.confirmed_at is not null then classes_users.class_id end) confirmed_class_count