實驗環境:

create table t (c int);
insert into t values
(15),(7),(9),(10),(7),(8),(20),(16),(9),(19),
(14),(10),(11),(10),(10),(12),(7),(10),(7),(9);
commit;

c 等於10,7,9的數量有 11個，超過了半數。
超過半數的 item資訊。

通用方式,不依賴具體資料庫。

select t5.*,t6.*,ifnull(round(t5.sumtotal/t6.total,2),0) result from (
select t3.rn,t3.total,t3.c,sum(ifnull(t4.total,0)) sumtotal from (
select
t1.*,
(
select count(case when t2.total>t1.total then 1 when t2.total=t1.total and t2.c<t1.c then 1 else null end)+1 from (
select c,count(*) total from t group by c
) t2
) rn
from(
select c,count(*) total from t group by c order by total desc ,c
) t1
) t3
left join(
select
t1.*,
(
select count(case when t2.total>t1.total then 1 when t2.total=t1.total and t2.c<t1.c then 1 else null end)+1 from (
select c,count(*) total from t group by c
) t2
) rn
from(
select c,count(*) total from t group by c order by total desc ,c
) t1
) t4 on (t3.rn> t4.rn)
group by t3.rn,t3.total,t3.c
) t5
left join(
select count(*) total from t
) t6 on(1=1)
where ifnull(round(t5.sumtotal/t6.total,2),0)<=0.5;

查詢結果:

繫結MySQL的實現

select * from (
select t3.*,case when result<=0.5 then 1 when result>0.5 and @b=-1 then @b:=1 else null end r from (
select t1.*,round((@a:=@a+t1.total)/t2.total,2) result from (
select c,count(*) total from t,(select @a:=0,@b:=-1) vars group by c order by 2 desc
) t1
left join(
select count(*) total from t
) t2 on(1=1)
) t3 order by result
) t4 where r=1;

結果:

查詢前50%的資料

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