我需要根据以下表格计算花费的总时间：
id | start_time | end_time |
期间可以重叠。我只需要计算一次ovelpapping周期。
E.g。如果我有这样的时期：
*----A----* *------C-----* *----------D----------*
*-----B-----* *---E---*
总和将是：(A.end-A.start) + (C.end - B.start) + (D.end - D.start)
我对编写此查询时应该使用的方法感到困惑，并且非常感谢您的帮助。
答案 0 :(得分：2)
好的，我认真地坚持要你在生产中使用它之前以各种方式测试它。 特别是测试如果在1个时间跨度内存在MULTIPLE重叠会发生什么。
此查询的作用是计算每个时间跨度的持续时间，以及与具有更高ID的其他时间跨度存在多少重叠。
select
t1.id,
t1.start_time,
t1.end_time,
t1.end_time - t1.start_time as duration,
sum(
if(t2.start_time < t1.start_time and t2.end_time > t1.end_time , t1.end_time - t1.start_time, 0) -- t2 completely around t1
+ if(t2.start_time >= t1.start_time and t2.end_time <= t1.end_time , t2.end_time - t2.start_time, 0) -- t2 completely within t1
+ if(t2.start_time < t1.start_time and t2.end_time > t1.start_time and t2.end_time < t1.end_time , t2.end_time - t1.start_time, 0) -- t2 starts before t1 starts and overlaps partially
+ if(t2.start_time < t1.end_time and t2.end_time > t1.end_time and t2.start_time > t1.start_time, t1.end_time - t2.start_time, 0) -- t2 starts before t1 ends and overlaps partially
) as overlap
from
times t1
left join times t2 on
t2.id > t1.id -- t2.id is greater than t1.id
and (
(t2.start_time < t1.start_time and t2.end_time > t1.end_time ) -- t2 completely around t1
or (t2.start_time >= t1.start_time and t2.end_time <= t1.end_time ) -- t2 completely within t1
or (t2.start_time < t1.start_time and t2.end_time > t1.start_time) -- t2 starts before t1 starts and overlaps
or (t2.start_time < t1.end_time and t2.end_time > t1.end_time ) -- t2 starts before t1 ends and overlaps
)
group by
t1.id
所以你最终想要的是：
select
sum(t.duration) - sum(t.overlap) as filtered_duration
from
(
OTHER QUERY HERE
) as t
所以最后你有这个问题：
select
sum(t.duration) - sum(t.overlap) as filtered_duration
from
(
select
t1.id,
t1.start_time,
t1.end_time,
t1.end_time - t1.start_time as duration,
sum(
if(t2.start_time < t1.start_time and t2.end_time > t1.end_time , t1.end_time - t1.start_time, 0) -- t2 completely around t1
+ if(t2.start_time >= t1.start_time and t2.end_time <= t1.end_time , t2.end_time - t2.start_time, 0) -- t2 completely within t1
+ if(t2.start_time < t1.start_time and t2.end_time > t1.start_time and t2.end_time < t1.end_time , t2.end_time - t1.start_time, 0) -- t2 starts before t1 starts and overlaps partially
+ if(t2.start_time < t1.end_time and t2.end_time > t1.end_time and t2.start_time > t1.start_time, t1.end_time - t2.start_time, 0) -- t2 starts before t1 ends and overlaps partially
) as overlap
from
times t1
left join times t2 on
t2.id > t1.id -- t2.id is greater than t1.id
and (
(t2.start_time < t1.start_time and t2.end_time > t1.end_time ) -- t2 completely around t1
or (t2.start_time >= t1.start_time and t2.end_time <= t1.end_time ) -- t2 completely within t1
or (t2.start_time < t1.start_time and t2.end_time > t1.start_time) -- t2 starts before t1 starts and overlaps
or (t2.start_time < t1.end_time and t2.end_time > t1.end_time ) -- t2 starts before t1 ends and overlaps
)
group by
t1.id
) as t
答案 1 :(得分：2)
我想建议另一种方法来获得时间，同时确保结果是正确的。但我不知道，如何完成MySQL的完成。
我将在以下时间重复使用上述示例 - 甚至可能存在第3级条目“F”：
1 3 7 12 13 (15 16) 20
|----A----| |------C-----| |----------D----------|
|-----B-----| |---E---|
5 9 14 17
|F|
查询按时间排序的所有时间戳的组合列表，并添加每个“操作”的类型
SELECT 1 as onoff, start_time as time FROM table
UNION
SELECT -1 as onoff, end_time as time FROM table
ORDER BY time
使用临时计数器循环处理列表（？），在启动/登录时递增1，在结束/注销时递减1
计数器应该使脚本在tmp.start=<time>的临时表中添加新行，如果它从0更改为1
并更新temp中上一行的tmp.end=<time>。表，如果它从1变为0.
脚本会为上面的示例执行此操作，如下所示：
QUERY TMP TABLE
onoff | time | ctr ID | start | end
1 | 01:00 | 1 1 | 01:00 | (record 1 added, ctr 0->1)
-1 | 03:00 | 0 1 | 01:00 | 03:00 (record 1 updated, ctr 1->0)
1 | 05:00 | 1 2 | 05:00 | (record 2 added, ctr 0->1)
1 | 07:00 | 2 (nothing to do)
-1 | 09:00 | 1 (nothing to do)
-1 | 12:00 | 0 2 | 05:00 | 12:00 (record 2 updated, ctr 1->0)
1 | 13:00 | 1 3 | 13:00 | (record 3 added, ctr 0->1)
1 | 14:00 | 2 (nothing to do)
1 | 15:00 | 3 (nothing to do)
-1 | 16:00 | 2 (nothing to do)
-1 | 17:00 | 1 (nothing to do)
-1 | 20:00 | 0 3 | 13:00 | 20:00 (record 3 updated, ctr 1->0)
最后一步非常简单：从timestampdiff()到start获取单位中的end，您需要/喜欢它并进行任何进一步过滤或分组
例如：在其他地方使用数据
SELECT ID, start, end, timestampdiff(MINUTE, start, end) FROM tmp
或者例如：总结每个用户登录的工作时间/时间
SELECT user_id, SUM(timestampdiff(MINUTE, start, end)) FROM tmp GROUP BY user_id
我敢肯定，这将为任何级别的嵌套提供正确的持续时间，但是有人知道如何在MySQL中完成此操作吗？我也想用它。
祝你好运
PS：如果脚本以＆gt;结束，脚本也可能“关闭”上一个会话或抛出错误。如果计数器变为＆lt; 1并抛出错误任何时候都是0
答案 2 :(得分：1)
我为another question写了一个类似的查询，所以我想我会根据这个问题调整它，任何人都有兴趣。
SELECT SUM(a.end_time - a.start_time) total_duration
FROM (
SELECT MIN(g.start_time) start_time, MAX(g.end_time) end_time
FROM (
SELECT @group_id := @group_id + (@end_time IS NULL OR o.start_time > @end_time) group_id,
start_time,
@end_time := CAST(CASE
WHEN (@end_time IS NULL OR o.start_time > @end_time) THEN o.end_time
ELSE GREATEST(o.end_time, @end_time)
END AS DATETIME) end_time
FROM times o
JOIN (SELECT @group_id := 0, @end_time := NULL) init
ORDER BY o.start_time ASC
) g
GROUP BY g.group_id
) a
最内层查询将您的时间组合在重叠组中，在适当的时候延伸end_time。 end_time用于处理前一个完全包含的时间。
下一个换行查询从每个组中提取整个时间范围。
外部查询总结了每个组的差异。