Question

我有一个看起来像这样的表：

updated_on          | start_date

2017-09-28 21:09:05 | 2017-08-06 06:59:04

2017-10-02 15:19:05 | 2017-09-28 21:11:21

2017-10-05 12:00:00 | 2017-10-02 15:21:05

我想做的是用start_date的第二个实例减去第一个更新的实例，意思是

TIMESTAMPDIFF(SECOND, 2017-09-28 21:09:05, 2017-09-28 21:11:21) 
TIMESTAMPDIFF(SECOND, 2017-10-02 15:19:05, 2017-10-02 15:21:05)

如果存在多行，

并继续这样做。

我如何实现这一目标？

Answer 1

您可以使用共同相关的子查询来获取下一条记录。

<强>查询

SELECT 
 *
 , (SELECT
     start_date 
   FROM 
     Table1 table12
   WHERE
     Table1.updated_on < table12.updated_on
   ORDER BY 
     updated_on ASC 
   LIMIT 1
  ) 
   as next_start_date
FROM
 Table1
ORDER BY
 updated_on ASC

结果（SQL小提琴）

|           updated_on |           start_date |      next_start_date |
|----------------------|----------------------|----------------------|
| 2017-09-28T21:09:05Z | 2017-08-06T06:59:04Z | 2017-09-28T21:11:21Z |
| 2017-10-02T15:19:05Z | 2017-09-28T21:11:21Z | 2017-10-02T15:21:05Z |
| 2017-10-05T12:00:00Z | 2017-10-02T15:21:05Z |               (null) |

demo http://www.sqlfiddle.com/#!9/94af3d/8

现在我们可以使用TIMESTAMPDIFF函数了。

<强>查询

SELECT 
 *
 , TIMESTAMPDIFF(
       SECOND 
     , updated_on
     , (SELECT
         start_date 
        FROM 
         Table1 table12
        WHERE
         Table1.updated_on < table12.updated_on
        ORDER BY 
         updated_on ASC 
        LIMIT 1
       )
   ) as diff

FROM
 Table1
ORDER BY
 updated_on ASC

<强>结果

|           updated_on |           start_date |   diff |
|----------------------|----------------------|--------|
| 2017-09-28T21:09:05Z | 2017-08-06T06:59:04Z |    136 |
| 2017-10-02T15:19:05Z | 2017-09-28T21:11:21Z |    120 |
| 2017-10-05T12:00:00Z | 2017-10-02T15:21:05Z | (null) |

demo http://www.sqlfiddle.com/#!9/94af3d/5

Answer 2

SELECT x.* 
     , TIMEDIFF(MIN(y.start_date),x.updated_on) n
  FROM my_table x 
  JOIN my_table y 
    ON y.updated_on > x.updated_on 
 GROUP 
     BY x.updated_on;

......或类似的东西

减去两行MySQL之间的时差

2 个答案: