在oralce SQL中，将行移到列，将列移到行

我从脚本中获得以下oracle SQL输出

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

我想使用PIVOT&UNPIVOT将行移动到列，将列移动到行，并在SQL中显示如下格式

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

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这很棘手；它需要一个解pivot和枢轴。这里有一种方法：

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;

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一个选项是使用条件聚合，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";

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要先取消数据绑定，然后通过条件聚合对数据进行整理。

@Gordan Linoff，谢谢您的快速回答。你知道如何使用unpivot和pivot实现这一点吗？Chris Saxon有一个关于这个主题的优秀博客。