Question

我需要将结果集旋转90度。我一直在玩数据透视和不旋转透视，但到目前为止还没有设法使任何东西起作用。结果集如下：

2018-07-03  TUE 20:00   12:15   465 (null)
2018-07-02  MON 20:00   20:05   -5  FAIL
2018-06-29  FRI 19:30   19:04   26  (null)
2018-06-28  THU 20:00   19:07   53  (null)
2018-06-27  WED 20:00   19:37   23  (null)

我需要的是能够将整个结果集旋转90度，以使第1列（日期）成为列标题，并且数据重置位于这些列的下方，例如：

2018-06-27  2018-06-28  2018-06-29  2018-07-02  2018-07-03
WED         THU         FRI         MON         TUE
20:00       20:00       19:30       20:00       20:00
19:37       19:07       19:04       20:05       12:15
23          53          56          -5          465
(null)      (null)      (null)      FAIL        (null)

是否存在另一个使之成为可能的功能，或者是我还没有想到的枢轴/非枢轴方法？

Answer 1

您可能会生成使用此SQL的逻辑

select vl as date1, myCol as myCol1
    from 
       (
        select to_char(myDate,'yyyy-mm-dd') as mD, 
               to_char(myDate,'DY') as Dy, 
               to_char(myVal) as mVl, t.* from tab t
         where to_char(mydate,'yyyy-mm-dd') = <a_date_parameter>
       )
        unpivot (vl for myCol in (mD,Dy,myHourStart, myHourEnd,mVl,status))

通过unpivot概念的贡献为：

create table tab( myDate date, myHourStart varchar2(5), myHourEnd varchar2(5), myVal int, status varchar2(15) );
insert all 
   into tab values( date'2018-07-03','20:00','12:15', 465, null )
   into tab values( date'2018-07-02','20:00','20:05', -5, 'FAIL' )
   into tab values( date'2018-06-29','19:30','19:04', 26,  null )
   into tab values( date'2018-06-28','20:00','19:07', 53, null )
   into tab values( date'2018-06-27','20:00','19:37', 23,  null ) 
select * from dual;

select date1, date2, date3, date4, date5 from
(  select vl as date1, myCol as myCol1
    from 
       (
        select to_char(myDate,'yyyy-mm-dd') as mD, 
               to_char(myDate,'DY') as Dy, 
               to_char(myVal) as mVl, t.* from tab t
         where to_char(mydate,'yyyy-mm-dd') = '2018-06-27'
       )
        unpivot (vl for myCol in (mD,Dy,myHourStart, myHourEnd,mVl,status)) 
         ) q1 join
  (  select vl as date2, myCol as myCol2
      from 
         (
          select to_char(myDate,'yyyy-mm-dd') as mD, 
                 to_char(myDate,'DY') as Dy,
                 to_char(myVal) as mVl, t.* from tab t
           where to_char(mydate,'yyyy-mm-dd') = '2018-06-28'
         )
          unpivot (vl for myCol in (mD,Dy,myHourStart, myHourEnd,mVl,status)) 
           ) q2 on ( q1.myCol1 = q2.myCol2  )  
join
  (  select vl as date3, myCol as myCol3
      from 
         (
          select to_char(myDate,'yyyy-mm-dd') as mD, 
                 to_char(myDate,'DY') as Dy,
                 to_char(myVal) as mVl, t.* from tab t
           where to_char(mydate,'yyyy-mm-dd') = '2018-06-29'
         )
          unpivot (vl for myCol in (mD,Dy,myHourStart, myHourEnd,mVl,status)) 
           ) q3 on ( q1.myCol1 = q3.myCol3  ) 
join
  (  select vl as date4, myCol as myCol4
      from 
         (
          select to_char(myDate,'yyyy-mm-dd') as mD, 
                 to_char(myDate,'DY') as Dy,
                 to_char(myVal) as mVl, t.* from tab t
           where to_char(mydate,'yyyy-mm-dd') = '2018-07-02'
         )
          unpivot (vl for myCol in (mD,Dy,myHourStart, myHourEnd,mVl,status)) 
           ) q4 on ( q1.myCol1 = q4.myCol4  )            
join
  (  select vl as date5, myCol as myCol5
      from 
         (
          select to_char(myDate,'yyyy-mm-dd') as mD, 
                 to_char(myDate,'DY') as Dy,
                 to_char(myVal) as mVl, t.* from tab t
           where to_char(mydate,'yyyy-mm-dd') = '2018-07-03'
         )
          unpivot (vl for myCol in (mD,Dy,myHourStart, myHourEnd,mVl,status)) 
           ) q5 on ( q1.myCol1 = q5.myCol5  );

SQL Fiddle Demo

Oracle 11g旋转结果表90度

1 个答案: