总数超过1的总和

Question

当前我执行以下代码

select  system,aid,mid,buysell,sdate,medate,namount,type, count(*)
from databasefile
where mid in ('123456')
group by  system,aid,mid,buysell,sdate,medate,namount,type
order by aid

我得到以下数据 '''

我要插入一个差异字段，该字段的计算方式为[（系统a的namount）-（系统z的namount）]。系统z的namount的计算方式为（购买的总namount-出售的总namount）。

请帮助我获得一个sql来实现这一目标，非常感谢，衷心感谢您的帮助。

Answer 1

计算为[（系统a的数量）-（系统z的数量）]

如果这是您想要的，则可以使用窗口功能：

select system, aid, mid, buysell, sdate, medate, namount, type, count(*),
       (sum(case when system = 'a' then namount else 0 end) over () -
        sum(case when system = 'z' then namount else 0 end) over ()
       )
from databasefile
where mid in ('123456')
group by system, aid, mid, buysell, sdate, medate, namount, type
order by aid