查找MySQL数据集的行之间的最小时间间隔

Question

我有一个MySQL表，记录了为用户收到的事件和收到事件的日期。

Date                    User

2012-01-21 18:30:02     AAA
2012-01-21 18:30:05     AAA
2012-01-21 18:30:08     AAA
2012-01-21 18:30:11     AAA
2012-01-21 18:30:15     AAA
2012-01-21 18:30:18     AAA
2012-01-21 18:30:21     AAA
2012-01-21 18:30:23     AAA
2012-01-21 18:30:26     AAA
2012-01-21 18:30:29     BBB
2012-01-21 18:30:32     BBB
2012-01-21 18:30:33     BBB
2012-01-21 18:30:37     BBB
2012-01-21 18:30:40     BBB
2012-01-21 18:30:42     BBB
2012-01-21 18:30:44     BBB
2012-01-21 18:31:01     BBB
2012-01-21 18:31:04     BBB
2012-01-21 18:31:07     BBB
2012-01-21 18:31:10     BBB

事件未按日期或用户排序。

我想找出单个用户的两个连续事件之间的时间间隔（以秒为单位）的最小值。所以结果集看起来像这样：

MIN_INTERVAL        USER

3                   AAA
5                   BBB

任何人都可以帮我提出一个生成此问题的SQL查询吗？我不认为GROUP BY会有所帮助。

Answer 1

这可以通过将自连接移动1个记录（按日期顺序）来实现，这将使得时间及其前一次的时间在同一行上，这样我们就可以在两者之间做出几秒的差异：

SELECT 
    a.user,
    MIN(TIMESTAMPDIFF(SECOND, b.date, a.date)) AS mindiff
FROM
(
    SELECT user, date, @val1:=@val1+1 AS rn
    FROM tbl
    CROSS JOIN (SELECT @val1:=0) AS val1_init
    ORDER BY date
) a
INNER JOIN
(
    SELECT user, date, @val2:=@val2+1 AS rn
    FROM tbl
    CROSS JOIN (SELECT @val2:=1) AS val2_init
    ORDER BY date
) b ON a.rn = b.rn
GROUP BY a.user

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SQLFiddle Demo

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Intermediate Results (Before GROUP BY)

Answer 2

简单版本可能不是最快的？

SELECT t1.user, MIN(TIMESTAMPDIFF(SECOND, t1.date, t2.date))
FROM tbl AS t1
JOIN tbl AS t2 ON t1.user = t2.user
WHERE t1.date < t2.date
GROUP BY t1.user

Answer 3

在mysql中，您需要使用自联接和两个分组来执行此操作：

select t.user, min(diff)
from (select t.user, unix_timestamp(tnext.date) - unix_timestamp(t.date) as diff
      from t join
           tnext
           on t.user = tnext.user and
              t.date < tnext.date
      group by t.user, t.date
     ) t
group by user

内部子查询下次查找并减去值以获取diff。外部返回最小的差异。