Question

我有一张包含以下2个相关字段的表格： -

QUOTEID  REG_DATE
001      02/10/2017
002      09/10/2017
003      10/10/2017
004      13/10/2017
005      13/10/2017
006      17/10/2017

是否可以让Oracle输出上述内容，但按星期（星期六到星期五）分组并计算，列名是相关星期五的日期？因此，使用上面的数据，输出将是： -

20/10/2017 13/10/2017 06/10/2017 
1          5          1

Answer 1

关于动态数据透视已经有很多问题，例如：Dynamic pivot in oracle sql。

此查询仅演示如何查找＆＃34;相关星期五＆＃34;如果您知道所有日期，则转动数据：

select * 
  from (select quoteid, 
               to_char(trunc(reg_date + 2, 'iw') + 4, 'dd/mm/yyyy') friday
          from t)
  pivot (count(quoteid) for friday in ('20/10/2017', '13/10/2017', '06/10/2017'))

测试：

with t(quoteid, reg_date) as (
    select '001', date '2017-10-02' from dual union all
    select '002', date '2017-10-09' from dual union all
    select '003', date '2017-10-10' from dual union all
    select '004', date '2017-10-13' from dual union all
    select '005', date '2017-10-13' from dual union all
    select '006', date '2017-10-17' from dual)
select * 
  from (select quoteid, 
               to_char(trunc(reg_date + 2, 'iw') + 4, 'dd/mm/yyyy') friday
          from t)
  pivot (count(quoteid) for friday in ('20/10/2017', '13/10/2017', '06/10/2017'))

结果：

'20/10/2017' '13/10/2017' '06/10/2017'
------------ ------------ ------------
           1            4            1

根据日期动态返回计数列

1 个答案: