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

时间：2024-04-15 08:27:41

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.****.net/Seandba/article/details/105152412 

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

aa128bd9b772d921814e068c67d5e7f4

问题测试数据：

--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;

image2问题核心是求多条记录之间的并集操作 ，我写的sql如下，

--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;

09e02bd2c2d7e12249c24dc5380b754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：求17日的的通道开放时长

--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

image5sql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;

image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

优化：

上述代码全表扫描5次，效率垃圾，小强优化到一次扫描搞定了，代码

--第8的特征：下一条记录开始时间  大于  截止当前行的最大结束时间；那么就把这部分时间记下来，最后减掉即可
select (max(end_time) - min(begin_time)) * 24 -
       sum(decode(sign(next_begin_time - max_end_time),
                  1,
                  (next_begin_time - max_end_time) * 24,
                  0)) 通道开通时间
  from (select a.channel,
               a.begin_time,
               a.end_time,
               max(a.end_time) over(partition by a.channel order by a.begin_time rows between unbounded preceding and current row) max_end_time, --截止当前行的最大结束时间
               lead(a.begin_time, 1) over(partition by a.channel order by a.begin_time) next_begin_time --下一条记录的开始时间
          from xcp a) tmp;

我用plsql也优化了一个出来

/*思路：    
第一步：两两合并，两条记录之间的关系只有两种：有交集 和 无交集
        1)对于有交集的：两两合并，取MIN(begin_time),MAX(end_time)作为新记录，
        2)对于无交集的：同样取MIN(begin_time),MAX(end_time)作为新记录，不过把中间空白部分计入duration_del
第二步：然后将第一步合并的新纪录和下一条记录再两两合并，以此类推，直至合并完所有记录
第三步：结果就是 最终合并记录的  end_time-begin_time-duration_del*/
DECLARE
    duration_del NUMBER:=0;--存储无交集的两两记录之间的空白时间
    --用于存储合并后的时间
    begin_time_merge DATE; end_time_merge DATE;
    --用于输入要查询的时间段
    day1 DATE:=to_date(20200314,\'yyyymmdd\');
    day2 DATE:=to_date(20200330,\'yyyymmdd\');
BEGIN
    FOR i IN (SELECT ROWNUM rnow,aa.* FROM 
                          (SELECT a.channel,GREATEST(a.begin_time, day1) begin_time,LEAST(a.end_time,day2) end_time
                            FROM xcp a WHERE NOT (end_time < day1 OR  begin_time> day2) ORDER BY 2)aa
                    )LOOP  --扫描一次全表
        IF i.rnow=1 THEN   --第一条记录用于初始化begin_time_merge  end_time_merge
            begin_time_merge :=i.begin_time; end_time_merge:=i.end_time;
        ELSE 
             IF i.begin_time>end_time_merge THEN
                duration_del:= duration_del+ (i.begin_time-end_time_merge)*24;--空白部分计入duration_del
             END IF;
             end_time_merge := GREATEST(end_time_merge,i.end_time); 
        END IF;        
    END LOOP;  
     DBMS_OUTPUT.PUT_LINE((end_time_merge-begin_time_merge)*24-duration_del||\'个小时通道开放\');
END;
/

就这样 以备后查

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