标签：计划 tle tmp splay 节点 height article sign block
来自于一个基友的问题：
他的博客同问题链接 sql时间段取并集、合并 https://blog.csdn.net/Seandba/article/details/105152412
问题：计算通道的总开放时长，只要有任意一个终端开放通道就算开放，难点在于各种终端开放时间重叠包含
--问题一、测试数据--计算总开放时长（小时） TRUNCATE TABLE xcp; insert into xcp values(‘1‘,‘A1‘,to_date(‘20200317 01:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 06:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘2‘,‘A1‘,to_date(‘20200317 01:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 06:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘2‘,‘A1‘,to_date(‘20200317 01:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 08:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘2‘,‘A1‘,to_date(‘20200317 02:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 07:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘2‘,‘A1‘,to_date(‘20200317 03:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 07:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘2‘,‘A1‘,to_date(‘20200317 05:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 09:00:00‘,‘yyyymmdd hh24:mi:ss ‘)); insert into xcp values(‘3‘,‘A1‘,to_date(‘20200317 09:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 11:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘3‘,‘A1‘,to_date(‘20200317 12:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 13:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘2‘,‘A1‘,to_date(‘20200317 14:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 19:00:00‘,‘yyyymmdd hh24:mi:ss ‘)); insert into xcp values(‘3‘,‘A1‘,to_date(‘20200317 16:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 19:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘3‘,‘A1‘,to_date(‘20200317 18:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 19:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘3‘,‘A1‘,to_date(‘20200317 18:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 21:00:00‘,‘yyyymmdd hh24:mi:ss‘)); commit; SELECT * FROM xcp;
--问题1 WITH tmp1 AS ( --取所有时间节点 SELECT channel,BEGIN_TIME TIME FROM xcp UNION SELECT channel,end_time FROM xcp UNION SELECT channel,MIN(begin_time) FROM xcp GROUP BY channel UNION SELECT channel,MAX(end_time) FROM xcp GROUP BY channel), tmp2 AS(--每个时间节点连接到下个节点 形成时间段 SELECT a.channel,a.time,LEAD(a.time,1) OVER(PARTITION BY a.channel ORDER BY a.time) nexttime FROM tmp1 a), tmp3 AS(--每个时间段取中值 SELECT b.channel,b.TIME,b.nexttime,(b.nexttime-b.time)/2+b.time midtime FROM tmp2 b WHERE b.nexttime IS NOT NULL), tmp4 AS(--若中值处于原始记录中 则该段时间为通道开通时间 否则通道不开通 SELECT c.*, CASE WHEN EXISTS (SELECT 1 FROM xcp o WHERE c.midtime BETWEEN o.begin_time AND o.end_time) THEN 1 ELSE 0 END * (c.nexttime-c.time)*24 duration FROM tmp3 c) SELECT nvl(d.channel,‘合计时长‘) 通道,d.TIME 开始时间,d.nexttime 结束时间, SUM(duration) "通道开通时间(小时)" FROM tmp4 d GROUP BY rollup((d.channel,d.TIME,d.nexttime)) ORDER BY 2;
看着就很垃圾的sql，执行计划一定垃圾，记录以备后查询吧
原理是吧时间节点拿出来，对没两个时间节点之间的时间段，取中间值到原始记录表查询，如果是，这段时间就是属于并集后的，然后对并集后的记录求和
--问题2、测试数据--计算27号开放时长（小时） TRUNCATE TABLE xcp; insert into xcp values(‘13‘,‘A1‘,to_date(‘20200314 08:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200315 09:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘14‘,‘A1‘,to_date(‘20200317 08:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 09:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘15‘,‘A1‘,to_date(‘20200316 03:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200317 05:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘16‘,‘A1‘,to_date(‘20200317 08:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200318 10:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘17‘,‘A1‘,to_date(‘20200316 08:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200318 10:00:00‘,‘yyyymmdd hh24:mi:ss‘)); insert into xcp values(‘18‘,‘A1‘,to_date(‘20200320 08:00:00‘,‘yyyymmdd hh24:mi:ss‘),to_date(‘20200321 10:00:00‘,‘yyyymmdd hh24:mi:ss‘)); commit; SELECT * FROM xcp ORDER BY begin_time
sql如下：
----问题2 WITH tmp1 AS ( --取所有时间节点 取17号就加入17号0点和24点两个时间 SELECT channel,BEGIN_TIME TIME FROM xcp UNION SELECT channel,end_time FROM xcp UNION SELECT channel,MIN(begin_time) FROM xcp GROUP BY channel UNION SELECT channel,MAX(end_time) FROM xcp GROUP BY channel UNION SELECT DISTINCT channel,to_date(‘20200317‘,‘yyyymmdd‘) FROM xcp UNION SELECT DISTINCT channel,to_date(‘20200318‘,‘yyyymmdd‘) FROM xcp), tmp2 AS(--每个时间节点连接到下个节点 形成时间段 SELECT a.channel,a.time,LEAD(a.time,1) OVER(PARTITION BY a.channel ORDER BY a.time) nexttime FROM tmp1 a), tmp3 AS(--每个时间段取中值 SELECT b.channel,b.TIME,b.nexttime,(b.nexttime-b.time)/2+b.time midtime FROM tmp2 b WHERE b.nexttime IS NOT NULL AND to_char(b.TIME,‘yyyymmdd‘)=20200317), tmp4 AS(--若中值处于原始记录中 则该段时间为通道开通时间 否则通道不开通 SELECT c.*, CASE WHEN EXISTS (SELECT 1 FROM xcp o WHERE c.midtime BETWEEN o.begin_time AND o.end_time) THEN 1 ELSE 0 END * (c.nexttime-c.time)*24 duration FROM tmp3 c) SELECT nvl(d.channel,‘合计时长‘) 通道,d.TIME 开始时间,d.nexttime 结束时间, SUM(duration) "通道开通时间(小时)" FROM tmp4 d GROUP BY rollup((d.channel,d.TIME,d.nexttime)) ORDER BY 2;
思路是在第一步取时间节点的时候单独加入17日0点24点的时间点即可
上述代码全表扫描5次，效率垃圾，从小强的第8种情况的反面考虑，结合小强给的思路，即可优化到扫描一次全表即可，代码如下
--第8的特征：下一条记录开始时间 大于 本条记录的结束时间；那么就把这部分时间记下来，最后减掉即可 WITH tmp AS( SELECT a.channel,a.begin_time,a.end_time, (LEAD(a.begin_time,1) OVER(PARTITION BY a.channel ORDER BY begin_time,end_time) - a.end_time)*24 hoursto_next_begin_time --距离下一条记录的时间间隔 如果是正数就是第8种情况 FROM xcp a) SELECT (MAX(end_time)-MIN(begin_time))*24 - sum(DECODE(sign(hoursto_next_begin_time),1,hoursto_next_begin_time,0)) 通道开通时间 FROM tmp aa
用sql对含有时间段字段(起始时间、结束时间)的记录做并集处理
标签：计划 tle tmp splay 节点 height article sign block
原文地址：https://www.cnblogs.com/yongestcat/p/12590154.html
