Question

我有以下三个表

tab_2016
+-----+------+---------+
|  id | month| salary  |
+-----+------+---------+
| 002 |  aug |  500    |
| 002 |  sep |  400    |
+-----+------+---------+


tab_2017    
+-----+------+---------+
|  id | month| salary  |
+-----+------+---------+
| 001 |  jan | 1000    |
| 001 |  jul | 2000    |
| 002 |  aug |  500    |
| 002 |  sep |  400    |
+-----+------+---------+

tab_2018    
+-----+------+---------+
|  id | month| salary  |
+-----+------+---------+
| 001 |  feb | 500     |
| 001 |  jul | 400     |
| 002 |  aug | 300     |
| 002 |  sep | 400     |
+-----+------+---------+

我试图按如下方式逐年获取id为001的总薪水。

    +-----+---------+---------+-----------+
    |  id | YR_2017 | YR_2018 |  YR_2016  | 
    +-----+---------+---------+-----------+
    | 001 |  3000   | 900     |    0      |
    +-----+---------+---------+-----------+

我曾尝试使用case语句进行以下查询，但无法获得理想的结果。我尝试过使用联接，但是当表数量增加时，根据年份动态形成查询变得更加棘手。

    select id,
    case 
    when YR=2017
    then
    sal end as YR_2017,
    case 
    when YR=2018
    then
    sal end as YR_2018,
case 
    when YR=2016
    then
    sal end as YR_2016

 from 
    (select id,sum(salary) as sal,"2017" as YR   from tab_2017 where id=001 group by id
    union all
    select id,sum(salary) as  sal,"2018" as YR from tab_2018 where id=001 group by id
     union all
select id,sum(salary) as  sal,"2016" as YR from tab_2016 where id=001 group by id
) as a

Answer 1

select a.id, a.YR_2017, b.YR_2018, If(c.YR_2016 IS NULL,0,c.YR_2016 ) from (select id,sum(salary) as YR_2017   from tab_2017 where id=1 group by id) a
JOIN (select id,sum(salary) as YR_2018 from tab_2018 where id=1 group by id) b
on a.id=b.id
LEFT JOIN (select id,sum(salary) as YR_2016 from tab_2016 where id=1 group by id) c
on b.id=c.id;

这是我尚未测试的粗略查询

在Impala中将列转换为行

1 个答案: