MYSQL

Question

所以这是 table

ID     Date                 Event
1547   2013-05-09 16:26:02  PA
1547   2013-05-15 13:59:23  TA
1547   2013-05-21 10:16:56  EA
1547   2013-05-21 10:17:27  PM
1547   2014-01-16 11:42:12  IH
1547   2014-01-16 11:42:13  RP
1547   2014-01-21 10:01:18  MP <-
1547   2014-12-05 14:32:21  RE <-
1547   2014-12-05 14:34:24  RE
1666   2013-05-29 11:26:38  PA
1666   2013-06-04 13:38:42  TA
1666   2013-06-05 14:16:13  EA
1666   2013-08-21 10:07:08  PR
1666   2013-08-21 10:38:51  TR
1666   2013-08-21 10:38:52  MP <-
1666   2013-10-07 16:26:46  PM <-
1666   2013-10-09 14:38:51  TM
1666   2013-10-09 14:38:52  EP
1666   2013-10-25 10:29:01  IH
1666   2013-12-13 08:52:41  IH
1666   2013-12-13 08:52:43  RP
1666   2014-01-21 09:55:10  MP <-
1666   2014-05-05 15:52:34  AB <-
1666   2014-05-07 14:55:58  RD
1692   2013-06-10 14:17:17  PA
1692   2013-06-10 14:17:53  TA
1692   2013-06-10 15:01:08  EA
1692   2013-08-21 10:04:39  PR
1692   2013-08-21 10:37:38  TR
1692   2013-08-21 10:37:39  ER
1692   2013-09-26 08:33:48  PM
1692   2013-09-26 16:32:46  TM
1692   2013-09-26 16:32:47  EP
1692   2013-10-11 09:21:26  IH
1692   2013-12-19 15:29:20  IV
1692   2013-12-19 15:29:21  RP
1692   2013-12-19 15:33:19  MP <-
1692   2014-01-21 11:05:46  FX <-
1692   2014-01-22 10:16:27  RE

我想 select 每个 ID 的第一个“MP”事件，下一个立即发生并计算天数差异。

结果应该是：

ID     Date                 DaystoNextEVENT
1547   2014-01-21 10:01:18  318
1666   2013-08-21 10:38:52  47
1666   2014-01-21 09:55:10  104
1692   2013-12-19 15:33:19  33

我曾尝试使用限制，但总是无法限制具有匹配日期来源的同一 ID 的下一行。

Answer 1

对于版本 8+：

SELECT id,
       `date`,
       DATEDIFF(LEAD(`date`) OVER (PARTITION BY id ORDER BY `date`),  -- next date
                `date`                                                -- the date of MP
                ) DaystoNextEVENT
FROM table t1
WHERE event = 'MP'
  AND NOT EXISTS ( SELECT NULL                 -- check that there is no MP earlier
                   FROM table t2
                   WHERE t1.id = t2.id
                     AND t1.`date` > t2.`date`
                     AND t2.event = 'MP' );

版本 5.x:

SELECT t1.id,
       t1.`date`,
       DATEDIFF(t2.`date`, t1.`date`) DaystoNextEVENT
FROM table t1                                  -- row with MP
JOIN table t2 ON t1.id = t2.id                 -- next row
             AND t1.`date` < t2.`date`
WHERE t1.event = 'MP'
  AND NOT EXISTS ( SELECT NULL                 -- check that there is no MP earlier
                   FROM table t3
                   WHERE t1.id = t3.id
                     AND t1.`date` > t3.`date`
                     AND t3.event = 'MP' )
  AND NOT EXISTS ( SELECT NULL                 -- check that there is no a row
                   FROM table t4               -- between rows from tables t1 and t2
                   WHERE t1.id = t4.id
                     AND t1.`date` < t4.`date`
                     AND t4.`date` < t2.`date` );

Answer 2

我们可以在 sub-query 中使用光标，使用 order by day desc 我们可以反转排序顺序，从而给出下一个事件的日期。如果它是相同的 ID，我们将在查询中 return 它。

SET @quot='';
SET @id=0;
select
  id,
  day,
  lag_day,
  event,
  case when id=lag_id then
     datediff(lag_day,day) end 
     as days_to_next
from
(select
  @id lag_id,
  @id:=id id,
  @quot lag_day, 
  @quot:=day day,
  event
from 
  events
order by 
  id , 
  day desc) e
where event ='MP'
order by
  id,
  day;

  id | day                 | lag_day             | event | days_to_next
---: | :------------------ | :------------------ | :---- | -----------:
1547 | 2014-01-21 10:01:18 | 2014-12-05 14:32:21 | MP    |          318
1666 | 2013-08-21 10:38:52 | 2013-10-07 16:26:46 | MP    |           47
1666 | 2014-01-21 09:55:10 | 2014-05-05 15:52:34 | MP    |          104
1692 | 2013-12-19 15:33:19 | 2014-01-21 11:05:46 | MP    |           33

db<>fiddle here

MYSQL - Select 行匹配事件代码并立即处理下一行并计算时间日期差异

MYSQL - Select rows matching event code and imediate next row and calculate time date difference

datediff