oracle复杂查询是sql

一、over()分析函数

  分组查前几条:select * from test t where (select count(*) from test a where t.type=a.type and t.scope>a.scope)<2;

  --rank()/dense_rank() over(partition by ...order by ...)
  select * from(select t.*,rank() over(partition by t.type order by t.scope ) a from TEST t) a  where a.a<3

--dense_rank()分级 连续排序
select t.*,dense_rank() over(partition by t.type order by t.scope)a from test t
--rank()分级 跳跃排序
select t.*,rank() over(partition by t.type order by t.scope)a from test t

select * from Test t where 2>(select count(*) from Test a where t.type=a.type and t.scope>a.scope)
select t.* from Test t,(select a.type,max(a.scope) scope from TEST a group by a.type) d  where t.type=d.type and t.scope=d.scope

--笛卡尔乘积
select * from Test t,Test a
select t.* from Test t,(select a.type,max(a.scope) maxscope,min(a.scope) minscope from TEST a group by a.type) d  where t.type=d.type and t.scope=d.scope


  --
select t.*,d.maxscope-t.scope maxscope,t.scope-d.minscope minscope
  from Test t,
       (select a.type, max(a.scope) maxscope, min(a.scope) minscope
          from TEST a
         group by a.type) d
 where t.type = d.type

--min()/max() over(partition by ...)
select t.*,
       nvl(max(t.scope) over(partition by t.type), 0) - t.scope maxscope,
       t.scope - nvl(min(t.scope) over(partition by t.type), 0) minscope
  from test t

--lead()/lag() over(partition by ... order by ...)  
select t.*,lead(t.scope,1,0)over(partition by t.type order by t.scope) a--同组后一个
       from test t
select t.*,lag(t.scope,1,0)over(partition by t.type order by t.scope) a--同组前一个
       from test t



select t.*,
       first_value(t.scope) over(partition by t.type) first_sal,
       last_value(t.scope) over(partition by t.type) last_sal,
       sum(t.scope) over(partition by t.type) sum_sal,
       avg(t.scope) over(partition by t.type) avg_sal,
       count(t.scope) over(partition by t.type) count_num,
       row_number() over(partition by t.type order by t.scope) row_num
  from test t


--注:带order by子句的方法说明在使用该方法的时候必须要带order by


原文地址:https://www.cnblogs.com/bigmonkeys/p/7896323.html