oracle - 根据值对列重新排序

Question

有没有办法根据值对表（或查询）中的列进行重新排序。

例如，在每一行上，FacName 应该是第一个，NPI - 2nd，TIN - 3rd 和 Address - 4th。

谢谢你！

Answer 1

select col1, col2, col3, col4 from (
  select * from (
    select 
      n, str, row_number() over (partition by n order by 
      decode(substr(str,1,3),'Fac',1,'NPI',2,'TIN',3,'Add',4)) cn 
    from (
      select * from (select t.*, rownum n from t)
      unpivot (str for cn in (col1 as 0, col2 as 0, col3 as 0, col4 as 0))
    )
  )
  pivot (min(str) for cn in (1 as col1, 2 as col2, 3 as col3, 4 as col4))
)
order by n

小提琴

Answer 2

您想要在列之间移动数据这一事实强烈暗示底层数据模型已损坏并且需要进行规范化。修复数据模型比给系统增加另一层复杂性要合适得多。

话虽这么说，你应该能够做类似的事情

SELECT (CASE WHEN column1 LIKE 'FacilityName%' THEN column1
             WHEN column2 LIKE 'FacilityName%' THEN column2
             WHEN column3 LIKE 'FacilityName%' THEN column3
             WHEN column4 LIKE 'FacilityName%' THEN column4
             ELSE null
         END) column1,
       (CASE WHEN column1 LIKE 'NPI%' THEN column1
             WHEN column2 LIKE 'NPI%' THEN column2
             WHEN column3 LIKE 'NPI%' THEN column3
             WHEN column4 LIKE 'NPI%' THEN column4
             ELSE null
         END) column2,
       (CASE WHEN column1 LIKE 'TIN%' THEN column1
             WHEN column2 LIKE 'TIN%' THEN column2
             WHEN column3 LIKE 'TIN%' THEN column3
             WHEN column4 LIKE 'TIN%' THEN column4
             ELSE null
         END) column3,
       (CASE WHEN column1 LIKE 'Address%' THEN column1
             WHEN column2 LIKE 'Address%' THEN column2
             WHEN column3 LIKE 'Address%' THEN column3
             WHEN column4 LIKE 'Address%' THEN column4
             ELSE null
         END) column4
  FROM( <<your query>> )