Question

我从SQL查询中获得以下输出：

Project_Manager_ID   PROject_Rank  On_boarding_Gap    Project_Gap    End_Gap
101                   1                   2                 3           0.833
101                   2                   7                 1           3.4
101                   3                   0                 -3           0
101                   4                   3                 1            1.2
102                   1                   -1                 0            1
102                   2                    2                 3            2
102                   3                    5                 2           -3

这是业务规则：

当等级= 1时，则On_boarding_gap

当等级> 1和等级<（根据Project_Manager_ID的最大（等级）组）然后是Project_gap

当根据Project_Manager_ID排名=最大（排名）分组时，则End_Gap

添加所有这些不同的差距，并找到每个Project_Manager_ID的差距总值。忽略所有负值。因此，在上面的示例中， Project_Manager_ID = 101

On_boarding_gap = 2

Project_gap = 1＆-3 = 1（忽略-3值）

End_Gap = 1.2

总差距为（2 + 1 +1.2）= 4.2。

对于Project_Manager_ID = 102

On_boarding_gap = -1 = 0（忽略-1值）

Project_gap = 3

End_Gap = -3 = 0（忽略-3值）

总差距为（0+ 3 + 0）= 3

我不确定如何为此编写查询。

Answer 1

嗯。。。 case表达式和窗口函数可为您提供每一行的空白：

select t.*,
       (case when project_rank = 1
             then on_boarding_gap
             when project_rank < max(project_rank) over (partition by Project_Manager_ID)
             then Project_gap
             else End_Gap
        end) as imputed_gap
from t;

然后聚合：

select Project_Manager_ID, sum(imputed_gap)
from (select t.*,
             (case when project_rank = 1
                   then on_boarding_gap
                   when project_rank < max(project_rank) over (partition by Project_Manager_ID)
                   then Project_gap
                   else End_Gap
              end) as imputed_gap
      from t
     ) t
where imputed_gap >= 0
group by Project_Manager_ID

根据另一列的条件值添加不同的列值

1 个答案: