sql - 在oralce SQL中将行移动到列和列到行

Question

我从我的脚本中得到以下 oracle SQL 输出。

Status     Item1 Item2 Item3 Item4
Processed     22    12    10     0
Error         11    22     0     0
Unsent        10    11     0    22

我想将行移动到列和列到行，并使用 PIVOT 和 UNPIVOT 在 SQL 中显示为以下格式。

Items   Processed  Error  Unsent
Item1          22     11      10
Item2          12     22      11 
Item3          10      0       0
Item4           0      0      22

Answer 1

一种选择是使用条件聚合，包括unpivot表达式：

select items as "Items",
       max(case when status = 'Processed' then value end) as "Processed",
       max(case when status = 'Error' then value end) as "Error",
       max(case when status = 'Unsent' then value end) as "Unsent"
  from tab
unpivot (value for items in ( Item1, Item2, Item3, Item4 ))
  group by items
  order by "Items";

Demo

首先对数据进行反透视，然后通过条件聚合将它们整理起来。

Answer 2

这很棘手；它需要一个非枢轴和枢轴。这是一种方法：

select item,
       sum(case when status = 'Processed' then val end) as processed,
       sum(case when status = 'Error' then val end) as error,
       sum(case when status = 'Unsent' then val end) as unsent
from ((select 'Item1' as item, status, item1 as val from t
      ) union all
      (select 'Item2' as item, status, item2 from t
      ) union all
      (select 'Item3' as item, status, item3 from t
      ) 
     ) i
group by item;