用两套sql

用分析函数。。lag就是上一条记录，lead是下一条记录...

select 

tmp.ORDER_DATE,

tmp.PRODUCT_ID,

tmp.QTY,

tmp.上一天日期,

t2.qty as 上一天数量,

tmp.qty-nvl(t2.qty,0) as 相减的数量

from (

select 

t1.*,

lag(t1.order_date,1) over (order by order_date) as 上一天日期

 from (

select to_date('25-09-2021','dd-mm-yyyy') order_date,1000 product_id,100 qty from dual union all

select to_date('26-09-2021','dd-mm-yyyy'),2000,150 from dual union all

select to_date('27-09-2021','dd-mm-yyyy'),1000,160 from dual union all

select to_date('28-09-2021','dd-mm-yyyy'),2000,120 from dual union all

select to_date('29-09-2021','dd-mm-yyyy'),2000,200 from dual union all

select to_date('30-09-2021','dd-mm-yyyy'),1000,90 from dual) t1) tmp

left join  (

select to_date('25-09-2021','dd-mm-yyyy') order_date,1000 product_id,100 qty from dual union all

select to_date('26-09-2021','dd-mm-yyyy'),2000,150 from dual union all

select to_date('27-09-2021','dd-mm-yyyy'),1000,160 from dual union all

select to_date('28-09-2021','dd-mm-yyyy'),2000,120 from dual union all

select to_date('29-09-2021','dd-mm-yyyy'),2000,200 from dual union all

select to_date('30-09-2021','dd-mm-yyyy'),1000,90 from dual) t2 on tmp.上一天日期=t2.order_date

order by tmp.order_date