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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)
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原文地址：http://www.cnblogs.com/zhulongchao/p/4572697.html
