Question

有人知道这是否可行。我有这张桌子：

    +---------+--------+----------+---------------+
    | SOR_NUM | SOR_ID | SOR_COST | SOR_LINE_COST |
    +---------+--------+----------+---------------+
    |       7 |      1 |       10 |            35 |
    |       7 |      1 |       10 |            37 |
    |       7 |      1 |       10 |            39 |
    |       7 |      2 |       20 |            35 |
    |       7 |      2 |       20 |            32 |
    +---------+--------+----------+---------------+

我想得到小计（可能有分组集），如下所示：

+---------+---------------+----------+---------------+
| SOR_NUM |    SOR_ID     | SOR_COST | SOR_LINE_COST |
+---------+---------------+----------+---------------+
|       7 | 1             |       10 |            35 |
|       7 | 1             |       10 |            37 |
|       7 | 1             |       10 |            39 |
|       7 | Total SOR_ID  |       10 |           111 |
|       7 | 2             |       20 |            35 |
|       7 | 2             |       20 |            32 |
|       7 | Total SOR_ID  |       20 |            67 |
|       7 | Total SOR_NUM |       30 |           178 |
+---------+---------------+----------+---------------+

Answer 1

您可以查看Oracle SQL functions for Aggregation的ROLLUP和GROUPING函数。

下面的查询给出了您想要的输出，但是会发生一些SOR_COST的调整。我将此归因于您的数据未充分规范化的事实。

select sub2.sor_num,
       case when sub2.sor_id is null then 'Total SOR_ID'
            when sub2.sor_cost is null then 'Total SOR_NUM'
            else to_char(sub2.sor_id) end sor_id,
       nvl(sub2.sor_cost, sub2.sum_sub1_one_sor_cost) sor_cost,
       nvl(sub2.sor_line_cost, sub2.sum_sub2_sum_sor_line_cost) sor_line_cost
  from (select case when (grouping(sub1.sor_num) = 0 and
                          grouping(sub1.sor_id) = 0 and
                          grouping(sub1.sor_cost) = 0 and
                          grouping(sub1.sor_line_cost) = 0) then 1
                    when (grouping(sub1.sor_num) = 0 and
                          grouping(sub1.sor_id) = 1 and
                          grouping(sub1.sor_cost) = 1 and
                          grouping(sub1.sor_line_cost) = 1) then 2
                    else 0 end show_row,
               max(sub1.sor_id) ord,
               sub1.sor_num,
               sub1.sor_id,
               sub1.sor_cost,
               sum(sub1.one_sor_cost) sum_sub1_one_sor_cost,
               sub1.sor_line_cost,
               sum(sub1.sum_sor_line_cost) sum_sub2_sum_sor_line_cost
          from (select grouping(x.sor_num) grp_1,
                       grouping(x.sor_id) grp_2,
                       grouping(x.sor_cost) grp_3,
                       grouping(x.sor_line_cost) grp_4,
                       x.sor_num,
                       x.sor_id,
                       x.sor_cost,
                       nvl2(x.sor_cost, null, min(x.sor_cost)) one_sor_cost,
                       sum(x.sor_line_cost) sor_line_cost,
                       nvl2(x.sor_line_cost, null, sum(x.sor_line_cost)) sum_sor_line_cost
                  from (select 7  sor_num, 1  sor_id, 22 sor_cost, 35 sor_line_cost from dual union all
                        select 7  sor_num, 1  sor_id, 22 sor_cost, 37 sor_line_cost from dual union all
                        select 7  sor_num, 1  sor_id, 22 sor_cost, 39 sor_line_cost from dual union all
                        select 7  sor_num, 2  sor_id, 22 sor_cost, 35 sor_line_cost from dual union all
                        select 7  sor_num, 2  sor_id, 22 sor_cost, 32 sor_line_cost from dual) x
                 group by x.sor_num,
                          x.sor_id,
                          rollup(x.sor_cost, x.sor_line_cost)) sub1
         where ((sub1.grp_3 = 1 and sub1.grp_2 = 0 and sub1.grp_1 = 0) or
               (sub1.grp_1 = 0 and sub1.grp_2 = 0 and sub1.grp_3 = 0 and
               sub1.grp_4 = 0))
         group by rollup(sub1.sor_num,
                         sub1.sor_id,
                         sub1.sor_cost,
                         sub1.sor_line_cost)) sub2
 where sub2.show_row > 0
 order by sub2.sor_num, sub2.show_row, sub2.ord, sub2.sor_cost nulls last

PL / SQL分组集和SUM

1 个答案: