SQL - 组合分组依据

Question

试图弄清楚如何组合这两个查询，但正在努力解决。

1.

select ia.name, count(archived_i) from (
    select * from incident i
    where i.archived = true
) as archived_i
right join incident_action ia on archived_i.incident_action_id = ia.id
group by ia.name
order by ia.name;

产量

Detention   3
Expulsion   0
Warning     2

2.

select in_s.name, count(archived_i) from (
    select * from incident i
    where i.archived = true
) as archived_i
right join incident_severity as in_s on archived_i.incident_severity_id = in_s.id
group by in_s.name
order by in_s.name;

产量

High    0
Low     5

我想把这些结合起来形成一些东西

Detention   High   0
Detention   Low    3
Expulsion   High   0
Expulsion   Low    0
Warning     High   0
Warning     Low    2

什么是正确、高效的方法？

Answer 1

你可以cross join引用表severity和action来生成所有可能的组合，然后把事件表带上left join。最后一步是聚合。

select ina.name action, ins.name severity, count(i.archived_i)
from incident_severity ins
cross join incident_action ina
left join incident i 
    on  i.incident_action_id = ina.id
    and i.incident_severity_id = ins.id
    and i.archived
group by ina.name, ins.name

Answer 2

在不更改查询的情况下，您可以进行交叉连接。例如：

select
  x.name,
  y.name,
  'x'
from (
  select ia.name, count(archived_i) from (
    select * from incident i
    where i.archived = true
  ) as archived_i
  right join incident_action ia on archived_i.incident_action_id = ia.id
  group by ia.name
  order by ia.name
) x
cross join (
  select in_s.name, count(archived_i) from (
    select * from incident i
    where i.archived = true
  ) as archived_i
  right join incident_severity as in_s 
    on archived_i.incident_severity_id = in_s.id
  group by in_s.name
  order by in_s.name
) y