按条件将列拆分为多列

Question

我有如下查询：

select
  table.date,
  table.shift,
  sum(table.value)
from
  db.table
where
  table.date >= date '2020-01-01' and
  table.filter = 'type'
group by
  table.date,
  table.shift
order by
  table.date,
  table.shift;

以这种方式返回数据：

date       | shift | sum(value)
-----------|-------|------------
2020-01-06 | 1     | 15
2020-01-06 | 3     | 12
2020-01-07 | 1     | 20
2020-01-07 | 2     | 38
2020-01-09 | 1     |  6
2020-01-09 | 2     | 22
2020-01-09 | 3     | 14
2020-01-10 | 1     | 17
2020-01-10 | 2     |  3
2020-01-10 | 3     | 10

我正在尝试这样获得，但我不知道如何：

date       | 1  | 2  | 3
-----------|----|----|----
2020-01-06 | 15 |    | 12
2020-01-07 | 20 | 38 |
2020-01-09 |  6 | 22 | 14
2020-01-10 | 17 |  3 | 10

Answer 1

您可以使用条件聚合

with cte as
(
select
  table.date,
  table.shift,
  sum(table.value) as val
from
  db.table
where
  table.date >= date '2020-01-01' and
  table.filter = 'type'
group by
  table.date,
  table.shift
order by
  table.date,
  table.shift
)
select date, max(case when shift=1 then val end) as 1,
max(case when shift=1 then val end) as 2,
max(case when shift=1 then val end) as 3
from cte
group by date

Answer 2

不需要附加子查询或CTE。

您可以使用条件聚合来对数据集进行透视，并稍加修改您的查询：只需从shift子句中删除group by，然后在sum() s中实现条件逻辑：

select
  t.date,
  sum(case when t.shift = 1 then t.value end) shift1,
  sum(case when t.shift = 2 then t.value end) shift2,
  sum(case when t.shift = 3 then t.value end) shift3
from
  db.table t
where
  t.date >= date '2020-01-01' and
  t.filter = 'type'
group by t.date
order by t.date

Answer 3

您可以进行条件聚合：

select t.date,
       sum(case when table.shift = 1 then table.value else 0 end),
       sum(case when table.shift = 2 then table.value else 0 end),
       sum(case when table.shift = 3 then table.value else 0 end)
from db.table as t
where t.date >= date '2020-01-01' and
      t.filter = 'type'
group by t.date;