SQL:值的总和,但仅限于不同的ID - 条件求和？

Question

我有以下结构:

具有typ1和typ2的多个事件的日,其中typ1和typ2具有各自日期的外键.Typ2也有持续时间.

现在我想计算所有typ1事件,所有typ2事件和typ2持续时间的总和.

示例数据:

天:

ID = 1 | Date = yesterday | ...

TYP1:

ID = 1 | FK_DAY = 1 | ...

ID = 2 | FK_DAY = 1 | ...

TYP2:

ID = 1 | FK_DAY = 1 | duration = 10

ID = 2 | FK_DAY = 1 | duration = 20

我现在想要结果:

Day.ID = 1 | countTyp1 = 2 | countTyp2 = 2 | sumDurationTyp2 = 30

我的问题是总和,我需要像"sum for distinct typ2.ID"这样的东西......有谁知道解决这个问题的方法？

我正在使用类似下面的东西,但那当然不能按我想要的方式工作:

SELECT day.id,
   count( DISTINCT typ1.id ),
   count( DISTINCT typ2.id ),
   sum( duration ) AS duration
FROM days
   LEFT JOIN typ
          ON day.id = typ1.id
   LEFT JOIN typ2
          ON day.id = typ2.id
GROUP BY day.id;

Answer 1

我的一般方法是在加入之前预先聚合每个表.

部分是因为你实际上并没有总结不同的值(如果两行中的每一行都有10,答案仍然是20).

但主要是因为它实际上更简单.子查询进行聚合,然后连接都是1:1.

SELECT
  days.id,
  typ_agg.rows,
  type2_agg.rows,
  type2_agg.duration
FROM
  days
LEFT JOIN
  (SELECT fk_day, COUNT(*) as rows FROM typ GROUP BY fk_day)  AS typ_agg
    ON days.id = typ_agg.fk_day
LEFT JOIN
  (SELECT fk_day, COUNT(*) as rows, SUM(duration) as duration FROM typ2 GROUP BY fk_day)  AS typ2_agg
    ON days.id = typ2_agg.fk_day