Session重叠问题学习(三)--优化

接前文
http://blog.itpub.net/29254281/viewspace-2150229/

前文中的算法想了一天半,终于在昨天晚上得出了正确的结果.
在我的环境中，耗时90s ，还有进一步优化的空间.

首选是生成 t1 和 t2的方式.
之前使用create table 方式导致类型不对,
因为是临时作用的表,所以可以预先创建表结构
CREATE TABLE `t1` (
`roomid` int(11) NOT NULL DEFAULT '0',
`userid` bigint(20) NOT NULL DEFAULT '0',
`s` timestamp ,
`e` timestamp,
primary KEY (`roomid`,`userid`,`s`,`e`),
KEY (`roomid`,`s`,`e`)
) ;

CREATE TABLE `t2` (
`roomid` int(11) NOT NULL DEFAULT '0',
`userid` bigint(20) NOT NULL DEFAULT '0',
`s` timestamp ,
`e` timestamp,
primary KEY (`roomid`,`userid`,`s`,`e`),
KEY (`roomid`,`s`,`e`)
) ;

前文中的第一步可以封装为一个过程

DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `p`()
BEGIN
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
;
END

函数修改如下

DELIMITER $$
CREATE DEFINER=`root`@`localhost` FUNCTION `f`(pTime timestamp) RETURNS int(11)
BEGIN
declare pResult bigint;
insert into t2
select distinct v6.roomid,v6.userid,greatest(s,starttime) s,least(e,endtime) e
from (
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!=''
) 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)
;
select row_count() into pResult;
RETURN pResult;
END

原来是针对每天每个房间处理,经过优化对某天的所有房间进行处理,批量的形式更快

另外在中间过程增加了类型转换,可以更好的利用索引
select roomid,CAST(starttime as DATETIME) starttime,CAST(endtime as DATETIME) endtime

另外第7行原来没有 distinct 可能导致bug
select distinct v6.roomid,v6.userid,greatest(s,starttime) s,least(e,endtime) e

调用时执行:
truncate table t1;
truncate table t2;
call p;
select f(s) from (
select distinct date(s) s from t1
) t

两步的执行时间：
Session重叠问题学习(三)--优化

今天优化了一天，从90s优化到25s以内,已经达到了预期。
我觉得在单线程环境,基本上已经达到最优.
如果还想优化到极致,第二步的函数执行,可以通过JAVA程序多线程一起跑,只要服务器CPU核数多,优化效果应该还是很明显的。

文章题目：Session重叠问题学习(三)--优化
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重庆分公司，新征程启航

Session重叠问题学习(三)--优化

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