Oracle SQL查询将十行一列转换为五行一列

我有一个以列名为值的表。表格中有10行。此输出的所需输出将显示为两列，前5行为一列A，第6至10行为一列B，彼此相邻为5行数据，如下所示

A  B
------------------------
ABN001001   ABN001006
ABN001002   ABN001007
ABN001003   ABN001008
ABN001004   ABN001009
ABN001005   ABN001010

数据脚本

create table Test1
(id number, value varchar(40));

begin
insert into Test1 (id,value) values (1000,'ABN001001');
insert into Test1 (id,value) values (1002,'ABN001002');
insert into Test1 (id,value) values (1003,'ABN001003');
insert into Test1 (id,value) values (1004,'ABN001004');
insert into Test1 (id,value) values (1005,'ABN001005');
insert into Test1 (id,value) values (1006,'ABN001006');
insert into Test1 (id,value) values (1007,'ABN001007');
insert into Test1 (id,value) values (1008,'ABN001008');
insert into Test1 (id,value) values (1009,'ABN001009');
insert into Test1 (id,value) values (1010,'ABN001010');
end;

谢谢
拉维

这是一种方法:

-为每一行分配一个行号
-通过将该数字除以5，将行分成1-5、6-10组。占据它的天花板
-同时获取此行号mod 5的值

然后，您可以按第一个计算中的值进行透视。这将它们分成两列。你需要两个mod将组分成五行。这是因为数据库为不在pivot子句中的每一列添加一个隐式组by

这给出了:

with rws as (
  select ceil ( row_number () over ( order by id ) / 5 ) grp,
         mod ( row_number () over ( order by id ), 5  ) rw,
         value
  from   test1 t
)
  select * 
  from   rws 
  pivot (
    max ( value ) for grp in ( 1 a, 2 b)
  )
  order by a;

RW   A           B           
   1 ABN001001   ABN001006   
   2 ABN001002   ABN001007   
   3 ABN001003   ABN001008   
   4 ABN001004   ABN001009   
   0 ABN001005   ABN001010