Oracle Sql优化之Merge 改写优化Update

时间:2021-09-28 18:14:50

1.待改写语句如下

update table1 f
set f.ljjine1= (select nvl(sum(nvl(b.jine1,0)),0) from table1 b where b.kjqj<=f.kjqj and b.gs=f.gs and b.bm=f.bm and b.yw=f.yw and b.currency=f.currency and substr(b.kjqj,1,4)=substr(f.kjqj,1,4)),
f.jine2 = (select nvl(sum(nvl(e.jine1,0)),0) from table2 e where e.kjqj=f.kjqj=e.gs=f.gs and e.bm=f.bm and e.yw= f.yw),
f.ljjine2 = (select nvl(sum(nvl(e.jine1,0)),0) from table2 e where e.kjqj<=f.kjqj and e.gs=f.gs and e.bm=f.bm and e.yw=f.yw and substr(e.kjqj,1,4)=substr(f.kjqj,1,4))
where substr(f.kjqj,1,4)= extract(year from sysdate)

2.分析语句:

a.第一个子查询除了等值条件,还有一个 “b.kjqj<=f.kjqj”非等值比较,因此这是一个累加,需要采用分析函数

b.第二个子查询有sum聚合函数,因此要把关联条件放入group by中,分组汇总

c.第三个子查询与第二个类似,只是等值条件改成不等值条件,所以要采用分析函数

3.子查询改写

第一个子查询改写如下

select b.rowid as rid ,sum(b.jine1) over (partition by b.gs,b.bm,b.yw,b.curreny order by b.kjqj) as ljjine1
from table1 b where substr(b.kjqj,1,4) = extract(year from sysdate)

第二个子查询改写,把关联列放到Select和group by后面

select e.gs,e.bm,e.yw,e.kjqj,sum(jine1) as jine2
from table2 e
where substr(e.kjqj,1,4)=extract(year from sysdate)
group by e.gs,e.bm,e.yw,e.kjqj

第三个子查询,可以在第二次子查询的基础上调用一次分析函数进行累加处理

select e.gs,e.bm,e.yw,e.kjqj,
sum(e.jine2) over(partition by e.gs,e.bm,e.yw order by e.kjqj) as ljjine2
from
(select e.gs,e.bm,e.yw,e.kjqj,sum(jine1) as jine2
from table2 e
where substr(e.kjqj,1,4)=extract(year from sysdate)
group by e.gs,e.bm,e.yw,e.kjqj) e

4.Merge改写的最终结果如下

Merge into table1 f
using(select b.rowid as rid ,sum(b.jine1) over (partition by b.gs,b.bm,b.yw,b.curreny order by b.kjqj) as ljjine1
from table1 b left join (
select e.gs,e.bm,e.yw,e.kjqj,
sum(e.jine2) over(partition by e.gs,e.bm,e.yw order by e.kjqj) as ljjine2
from
(select e.gs,e.bm,e.yw,e.kjqj,sum(jine1) as jine2
from table2 e
where substr(e.kjqj,1,4)=extract(year from sysdate)
group by e.gs,e.bm,e.yw,e.kjqj) e
) e on(b.gs=e.gs and b.bm=e.bm and b.yw=e.yw and b.kjqj=e.kjqj)
where substr(b.kjqj,1,4) = extract(year from sysdate)) b on (f.rowid=b.rid)
when matched then
update set
f.ljjine1= nvl(b.ljjine1,0),
f.ljjine2=nvl(b.ljjine2,0),
f.jine2 = nvl(b.jine2,,0)