行间时间差

Question

在 mysql 数据库 (ver.5.7.15) 中，我有一个名为 operation_list 的表，其中包含以下数据：id (autoincrement整数，主键），操作日期时间（日期时间），操作（枚举值：START,STOP），所以表格如下所示：

+------+---------------------+-----------+
| id   | operation_date_time | operation |
+------+---------------------+-----------+
|    1 | 2000-01-01 06:30:45 | START     |
|    2 | 2000-01-01 07:45:00 | STOP      |
|    3 | 2000-01-01 08:18:12 | START     |
|    4 | 2000-01-01 11:23:58 | STOP      |
|    5 | 2000-01-01 15:45:01 | START     |
|    6 | 2000-01-01 19:01:33 | STOP      |
+------+---------------------+-----------+

现在，假设第一行始终是 START，最后一行始终是 STOP，STOP 始终放在 START 之后，我需要检索 START 和 STOP 之间的时间差（以秒为单位）。因此，我需要编写一个 SQL 来生成以下记录集：

+------+---------------------+-----------+
| id   | operation_date_time | duration  |
+------+---------------------+-----------+
|    1 | 2000-01-01 06:30:45 | 4455      |
|    3 | 2000-01-01 08:78:12 | 11146     |
|    5 | 2000-01-01 15:45:01 | 11792     |  
+------+---------------------+-----------+

其中 4455 相当于 1 小时 14 分 15 秒， 11146 相当于 3 小时 5 分 46 秒， 11792 相当于 3 小时 16 分 32 秒，以此类推。

在不创建额外表或专用脚本的情况下，在单个 SQL 语句中执行此操作的最佳方法是什么？

Answer 1

这适用于 mysql 5.X

但它比 8.0 更丑

SELECT
   MIN(id) id
  ,MIN(`operation_date_time`) `operation_date_time`
  ,MAX(diff) duration  
FROM
(SELECT 
  id
  , IF(`operation` = 'START', 0,TIME_TO_SEC(TIMEDIFF(`operation_date_time`,  @datetime))) diff
  ,IF(`operation` = 'START',  @count := @count + 1,@count := @count) groupby
  ,@datetime := `operation_date_time` `operation_date_time`
 FROM
  (SELECT * FROM timetable ORDER by `operation_date_time` ASC) t1, (SELECT @datetime := NOW()) a,
   (SELECT @count := 0) b) t2
GROUP by groupby;

CREATE TABLE timetable (
  `id` INTEGER,
  `operation_date_time` VARCHAR(19),
  `operation` VARCHAR(5)
);

INSERT INTO timetable
  (`id`, `operation_date_time`, `operation`)
VALUES
  ('1', '2000-01-01 06:30:45', 'START'),
  ('2', '2000-01-01 07:45:00', 'STOP'),
  ('3', '2000-01-01 08:18:12', 'START'),
  ('4', '2000-01-01 11:23:58', 'STOP'),
  ('5', '2000-01-01 15:45:01', 'START'),
  ('6', '2000-01-01 19:01:33', 'STOP');

✓ ✓

SELECT
   MIN(id) id
  ,MIN(`operation_date_time`) `operation_date_time`
  ,MAX(diff) duration  
FROM
(SELECT 
  id
  , IF(`operation` = 'START', 0,TIME_TO_SEC(TIMEDIFF(`operation_date_time`,  @datetime))) diff
  ,IF(`operation` = 'START',  @count := @count + 1,@count := @count) groupby
  ,@datetime := `operation_date_time` `operation_date_time`
 FROM
  (SELECT * FROM timetable ORDER by `operation_date_time` ASC) t1, (SELECT @datetime := NOW()) a,
   (SELECT @count := 0) b) t2
GROUP by groupby;

编号 |操作日期时间 |期间 -: | :----------------- | --------: 1 | 2000-01-01 06:30:45 | 4455 3 | 2000-01-01 08:18:12 | 11146 5 | 2000-01-01 15:45:01 | 11792

db小提琴here

Answer 2

SELECT operation_date_time,DURATION FROM ( 从实践中选择 *,DATEDIFF(SECOND,operation_date_time,LEAD(operation_date_time)OVER(ORDER BY ID)) 作为持续时间 ）一个 WHERE 操作='START'

Answer 3

使用窗口函数获取结束时间。我将使用累积条件最小值：

select t.*,
       timestampdiff(second, operation_date_time, stop_dt) as diff_seconds
from (select t.*,
             min(case when operation = 'STOP' then operation_date_time end) over (order by operation_date_time) as stop_dt
      from t
     ) t
where operation = 'START';

如果数据真的是交错的，那么你可以使用lead():

select t.*,
       timestampdiff(second, operation_date_time, stop_dt) as diff_seconds
from (select t.*,
             lead(operation_date_time) over (order by operation_date_time) as stop_dt
      from t
     ) t
where operation = 'START';

编辑：

在 MySQL 8.0 之前的版本中：

select t.*,
       timestampdiff(second, operation_date_time, stop_dt) as diff_seconds
from (select t.*,
             (select min(t2.operation_date_time)
              from t t2
              where t2.operation = 'STOP' and
                    t2.operation_date_time > t.operation_date_time
             ) as stop_dt
      from t
     ) t
where operation = 'START';