SQL中上一行减去当前行

我有一张点菜的桌子

如何为传入列减去上一行减去当前行

在我的SQL中

 select a.Incoming, 
coalesce(a.Incoming - 
    (select b.Incoming from orders b where b.id = a.id + 1), a.Incoming) as differance
from orders a

在Oracle中使用函数

请尝试此查询

SELECT DATE_IN,Incoming,
   LAG(Incoming, 1, 0) OVER (ORDER BY Incoming) AS inc_previous,
   LAG(Incoming, 1, 0) OVER (ORDER BY Incoming) - Incoming AS Diff
FROM orders

LAG函数用于访问前一行的数据

SELECT DATE_IN,
       incoming_vol as incoming,
       LAG(incoming_vol, 1, incoming_vol) OVER (ORDER BY date_in) - incoming_vol AS incoming_diff
FROM orders
order by 1 desc

没有分析函数的另一种解决方案：

select o.date_in, o.incoming_vol, p.incoming_vol-o.incoming_vol
from orders p, orders o
where p.date_in = (select nvl(max(oo.date_in), o.date_in) 
                   from orders oo where oo.date_in < o.date_in)
;

---

上一行定义了什么？日期对于给定的样本数据，您的预期结果是什么？它是用于mysql还是oracle？试着使用lead oracle函数：Hi Abhik，我需要它作为列输入为什么使用PL/SQL标记？是否需要存储过程？按传入的顺序不正确。将前一行减去当前行，而不是请求下一行的差值。所以我认为，正确的方向，错误的答案-谢谢你是个救生员。我一直在努力解决这个问题

select o.date_in, o.incoming_vol, p.incoming_vol-o.incoming_vol
from orders p, orders o
where p.date_in = (select nvl(max(oo.date_in), o.date_in) 
                   from orders oo where oo.date_in < o.date_in)
;