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
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select roomid,as DATETIME) starttime,as DATETIME) endtime from (
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select @d as starttime,@d:=d,v3.roomid,v3.d endtime from (
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select distinct roomid,
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when nums.id=1 then v1s
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when nums.id=2 then v1e
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when nums.id=3 then v2s
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when nums.id=4 then v2e
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end d from (
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select v1.roomid, v1.s v1s,v1.e v1e,v2.s v2s,v2.e v2e
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from t1 v1
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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)
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where v2.roomid in(select distinct roomid from t1 where date(s)=pTime)
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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)
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) a,nums where nums.id<=4
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order by roomid,d
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) v3,(select @d:='') vars
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) 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.
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DELIMITER $$
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CREATE DEFINER=`root`@`localhost` PROCEDURE `p`()
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BEGIN
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declare done int default 0;
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declare v_roomid bigint;
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declare v_start timestamp;
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declare v_end timestamp;
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declare cur_test CURSOR for select roomid,s,e from t1 ;
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DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
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drop table if exists t1;
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drop table if exists tmp_time_point;
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CREATE temporary TABLE `t1` (
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`roomid` int(11) NOT NULL DEFAULT '0',
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`userid` bigint(20) NOT NULL DEFAULT '0',
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`s` timestamp NOT NULL DEFAULT ON UPDATE timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
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primary KEY `roomid` (`roomid`,`s`,`e`,`userid`)
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) ENGINE=InnoDB;
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create temporary table tmp_time_point(
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roomid bigint,
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timepoint timestamp,
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primary key(roomid,timepoint)
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) engine=memory;
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insert into t1
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select distinct
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roomid,
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userid,
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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
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from (
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SELECT DISTINCT s.roomid, s.userid, s.s, (
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SELECT MIN(e)
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FROM (SELECT DISTINCT roomid, userid, roomend AS e
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FROM u_room_log a
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WHERE NOT EXISTS (SELECT *
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FROM u_room_log b
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WHERE a.roomid = b.roomid
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AND a.userid = b.userid
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AND a.roomend >= b.roomstart
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AND a.roomend < b.roomend)
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) s2
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WHERE s2.e > s.s
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AND s.roomid = s2.roomid
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AND s.userid = s2.userid
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) AS e
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FROM (SELECT DISTINCT roomid, userid, roomstart AS s
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FROM u_room_log a
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WHERE NOT EXISTS (SELECT *
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FROM u_room_log b
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WHERE a.roomid = b.roomid
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AND a.userid = b.userid
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AND a.roomstart > b.roomstart
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AND a.roomstart <= b.roomend)
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) s, (SELECT DISTINCT roomid, userid, roomend AS e
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FROM u_room_log a
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WHERE NOT EXISTS (SELECT *
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FROM u_room_log b
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WHERE a.roomid = b.roomid
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AND a.userid = b.userid
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AND a.roomend >= b.roomstart
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AND a.roomend < b.roomend)
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) e
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WHERE s.roomid = e.roomid
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AND s.userid = e.userid
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) t1 ,
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nums
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where nums.id<=datediff(e,s)+1
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;
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open cur_test;
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repeat
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fetch cur_test into v_roomid, v_start,v_end;
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if done !=1 then
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insert ignore into tmp_time_point(roomid,timepoint) values(v_roomid,v_start);
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insert ignore into tmp_time_point(roomid,timepoint) values(v_roomid,v_end);
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end if;
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until done end repeat;
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close cur_test;
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select roomid,date(s) dt,round(second,s,e))/60) ts,max(c) c from (
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select roomid,s,e ,distinct userid) c from (
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select distinct v6.roomid,v6.userid,greatest(s,starttime) s,least(e,endtime) e
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from (
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select distinct roomid,as DATETIME) starttime,as DATETIME) endtime from (
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select
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if(@roomid=roomid,@d,'') as starttime,@d:=timepoint,@roomid:=roomid,p.roomid,p.timepoint endtime
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from tmp_time_point p,(select @d:='',@roomid:=-1) vars
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order by roomid,timepoint
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) v4 where starttime!='' and date(starttime)=date(endtime)
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) 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)
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) v6 group by roomid,s,e having distinct userid)>1
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) v7 group by roomid,date(s);
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END
都内聚到一个过程之后,不需要创建额外的普通表,直接在过程中创建临时表.实现高内聚,低耦合.
call p
过程返回的结果即为最终结果.
三次测试耗时均低于 10.3秒
当前名称:Session重叠问题学习(四)--再优化
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