Session重叠问题学习(四)--再优化

2024-04-02 19:55

短信预约 -IT技能 免费直播动态提醒

接前文:
需求描述和第一版解决方案(执行时间90秒)
http://blog.itpub.net/29254281/viewspace-2150229/

优化和修改bug的版本(执行时间25秒)
http://blog.itpub.net/29254281/viewspace-2150259/

我觉得在集合思维处理方式中,前文已经达到最优了.

如果放弃完全的集合处理思维,实际上还可以更加的优化.

前文的几个问题.
1.引入了过多的表结构.
2.写表本身也花费了时间.
3.前文按天批处理,粒度还是细了.应该一把批量全出最快.
4.前文计算最小间隔范围的部分,因为应用集合化思维,不好理解性能还差.

前文计算最小间隔范围的部分如下

select roomid,as DATETIME) starttime,as DATETIME) endtime from (
select @d as starttime,@d:=d,v3.roomid,v3.d endtime from (
select distinct roomid,
when nums.id=1 then v1s
when nums.id=2 then v1e
when nums.id=3 then v2s
when nums.id=4 then v2e
end d from (
select v1.roomid, v1.s v1s,v1.e v1e,v2.s v2s,v2.e v2e
from t1 v1
inner join t1 v2 on ((v1.s between v2.s and v2.e or v1.e between v2.s and v2.e ) and v1.roomid=v2.roomid)
where v2.roomid in(select distinct roomid from t1 where date(s)=pTime)
and v2.s>=pTime and v2.s<(pTime+interval '1' and (v2.roomid,v2.userid,v2.s,v2.e)!= (v1.roomid,v1.userid,v1.s,v1.e)
) a,nums where nums.id<=4
order by roomid,d
) v3,(select @d:='') vars
) v4 where starttime!=''

该部分使用集合处理方式,不好理解性能还差.

这块可以通过游标写临时表轻易解决。
本质上最小范围就是
每天每个房间每个记录的开始时间和结束时间都扣出来作为一行排序。

然后找到每个时间最近的下一个时间,作为最小时间范围.

如果使用游标,遍历一遍即可.

DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `p`()
BEGIN
declare done int default 0;
declare v_roomid bigint;
declare v_start timestamp;
declare v_end timestamp;
declare cur_test CURSOR for select roomid,s,e from t1 ;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
drop table if exists t1;
drop table if exists tmp_time_point;
CREATE temporary TABLE `t1` (
`roomid` int(11) NOT NULL DEFAULT '0',
`userid` bigint(20) NOT NULL DEFAULT '0',
`s` timestamp NOT NULL DEFAULT ON UPDATE timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
primary KEY `roomid` (`roomid`,`s`,`e`,`userid`)
) ENGINE=InnoDB;
create temporary table tmp_time_point(
roomid bigint,
timepoint timestamp,
primary key(roomid,timepoint)
) engine=memory;
insert into t1
select distinct
roomid,
userid,
if(date(s)!=date(e) and id>1,date(s+interval id-1 date(s+interval id-1 date(e) ,e,date_format(s+interval id-1 '%Y-%m-%d 23:59:59')) e
from (
SELECT DISTINCT s.roomid, s.userid, s.s, (
SELECT MIN(e)
FROM (SELECT DISTINCT roomid, userid, roomend AS e
FROM u_room_log a
WHERE NOT EXISTS (SELECT *
FROM u_room_log b
WHERE a.roomid = b.roomid
AND a.userid = b.userid
AND a.roomend >= b.roomstart
AND a.roomend < b.roomend)
) s2
WHERE s2.e > s.s
AND s.roomid = s2.roomid
AND s.userid = s2.userid
) AS e
FROM (SELECT DISTINCT roomid, userid, roomstart AS s
FROM u_room_log a
WHERE NOT EXISTS (SELECT *
FROM u_room_log b
WHERE a.roomid = b.roomid
AND a.userid = b.userid
AND a.roomstart > b.roomstart
AND a.roomstart <= b.roomend)
) s, (SELECT DISTINCT roomid, userid, roomend AS e
FROM u_room_log a
WHERE NOT EXISTS (SELECT *
FROM u_room_log b
WHERE a.roomid = b.roomid
AND a.userid = b.userid
AND a.roomend >= b.roomstart
AND a.roomend < b.roomend)
) e
WHERE s.roomid = e.roomid
AND s.userid = e.userid
) t1 ,
nums
where nums.id<=datediff(e,s)+1
;
open cur_test;
repeat
fetch cur_test into v_roomid, v_start,v_end;
if done !=1 then
insert ignore into tmp_time_point(roomid,timepoint) values(v_roomid,v_start);
insert ignore into tmp_time_point(roomid,timepoint) values(v_roomid,v_end);
end if;
until done end repeat;
close cur_test;
select roomid,date(s) dt,round(second,s,e))/60) ts,max(c) c from (
select roomid,s,e ,distinct userid) c from (
select distinct v6.roomid,v6.userid,greatest(s,starttime) s,least(e,endtime) e
from (
select distinct roomid,as DATETIME) starttime,as DATETIME) endtime from (
select
if(@roomid=roomid,@d,'') as starttime,@d:=timepoint,@roomid:=roomid,p.roomid,p.timepoint endtime
from tmp_time_point p,(select @d:='',@roomid:=-1) vars
order by roomid,timepoint
) v4 where starttime!='' and date(starttime)=date(endtime)
) v5 inner join t1 v6 on(v5.starttime between v6.s and v6.e and v5.endtime between v6.s and v6.e and v5.roomid=v6.roomid)
) v6 group by roomid,s,e having distinct userid)>1
) v7 group by roomid,date(s);
END