在 SQL (Oracle) 中将行划分为特定列

Question

一直在尝试一个场景，我想将返回数据分散到几列而不是冗长的行中

例如

COL1       |  COL2    |   COL3
-----------------------------
DATAHEAD   |  VALUE1  | VALUE9
DATAHEAD   |  VALUE2  | VALUE10 
DATAHEAD   |  VALUE3  | VALUE11
DATAHEAD   |  VALUE4  | VALUE12
DATAHEAD   |  VALUE5  | VALUE13
DATAHEAD   |  VALUE6  | VALUE14
DATAHEAD   |  VALUE7  | VALUE15
DATAHEAD   |  VALUE8  | VALUE16 

I want it to spread  into 6/n columns dynamically like 
this below

COL1       |  COL2     |   COL3   | COL4     | COL 5  |    COL6  |
--------------------------------------------------------------------
DATAHEAD   |  VALUE1   | VALUE2   |  VALUE3  | VALUE4  |  VALUE5  |
DATAHEAD   |  VALUE6   | VALUE7   | VALUE8   |  VALUE9 |  VALUE10 |
DATAHEAD   |  VALUE11  | VALUE12  |  VALUE13 | VALUE14 |  VALUE15 |
DATAHEAD   |  VALUE16  |  Null    | Null     |  Null   |  Null    |

数据数量并不重要，只要它像上面的格式一样分散即可。我看到了一些有关 Pivot() 的相关文章，但我找不到如何在这种情况下应用它。非常感谢您的帮助，因为我不太擅长 sql。

Answer 1

首先，我将排序后的单元格放入数据透视表中。

    select col1, col2 as col from table1
     union
    select col1, col3 as col from table1 
;

然后我提供了哪些行d和列m将是值的信息。

  select
    col1,
    col,
    trunc((rank() over (order by col)-1)/5) as d,
    mod(rank() over (order by col)-1,5) as m
  from
  (
    select col1, col2 as col from table1
     union
    select col1, col3 as col from table1 
  )
;

最后我根据文档做了一个pivot 。

select * from
(
  select
    col1,
    col,
    trunc((rank() over (order by col)-1)/5) as d,
    mod(rank() over (order by col)-1,5) as m
  from
  (
    select col1, col2 as col from table1
     union
    select col1, col3 as col from table1 
  )
)
pivot
(
  min(col) for m in (0 as COL2,1 as COL3,2 as COL4,3 as COL5,4 as COL6)
)
order by d
;

输出：

列1	D	列2	列3	彩色4	彩色5	列6
数据头	0	值01	值02	值03	值04	值05
数据头	1	值06	值07	值08	值09	值10
数据头	2	值11	值12	值13	值14	值15
数据头	3	值16	（无效的）	（无效的）	（无效的）	（无效的）

数据定义语言：

CREATE TABLE Table1
    ("COL1" varchar2(8), "COL2" varchar2(7), "COL3" varchar2(7))
;

INSERT ALL 
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE01', 'VALUE09')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE02', 'VALUE10')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE03', 'VALUE11')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE04', 'VALUE12')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE05', 'VALUE13')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE06', 'VALUE14')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE07', 'VALUE15')
    INTO Table1 ("COL1", "COL2", "COL3")
         VALUES ('DATAHEAD', 'VALUE08', 'VALUE16')
SELECT * FROM dual
;

编辑1：

Oracle中的rank over函数用于按顺序对行进行编号。

例如，rank() over (order by col)是一列，其中有从 1 开始的数字，如果按 col 排列该列，它将是 1, 2, 3, ... 如果重新排列行，排名仍将保留根据的值col 列。

因此，我之前的解决方案按字母顺序排列值，无论它们在源表中的位置如何。

如果您想按第一列排序并在最后按相同顺序对第二列排序，则可以使用此内部选择：

    select
      col1,
      rank() over (order by col2) as r,
      col2 as col
    from table1
    union
    select 
      col1,
      rank() over (order by col2) + (select count(*) from table1) as r,
      col3 as col
    from table1 
;

然后生成的选择将如下所示：

select * from
(
  select
    col1,
    col,
    trunc((r-1)/5) as d,
    mod(r-1,5) as m
  from
  (
    select
      col1,
      rank() over (order by col2) as r,
      col2 as col
    from table1
    union
    select 
      col1,
      rank() over (order by col2) + (select count(*) from table1) as r,
      col3 as col
    from table1 
  )
)
pivot
(
  min(col) for m in (0 as COL2,1 as COL3,2 as COL4,3 as COL5,4 as COL6)
)
order by d
;