Question

确定。我的SQL很漂亮，所以我很难解决这个问题。

我有一个表存储带有时间戳的记录。我想要的是uids的列表，其中在1秒的时间范围内，该用户有2条或更多条记录。也许我已经让它变得更加复杂，只是想不出来。

缩短版本的表格（pk忽略）

    uid date
    1   2015-01-01 10:00:30.020*
    1   2015-01-01 10:00:30.300*
    1   2015-01-01 10:00:30.500*
    1   2015-01-01 10:00:39.000
    1   2015-01-01 10:00:35.000
    1   2015-01-01 10:00:37.800
    2   2015-02-02 12:00:30.000
    2   2015-02-02 14:00:30.000
    2   2015-02-02 15:00:30.000
    2   2015-02-02 18:00:30.000 
    3   2015-03-02 15:00:24.000 
    3   2015-03-02 15:00:20.000 *
    3   2015-03-02 15:00:20.300 *

我已经标记了*我希望匹配的记录旁边。我喜欢的结果列表只是一个uid列表，所以我想要的结果就是 1 3

Answer 1

您可以使用exists：

执行此操作

select distinct uid
from t
where exists (select 1
              from t t2
              where t2.uid = t.uid and
                    t2.date > t.date and
                    t2.date <= t.date + interval 1 second
             );

注意：添加1秒的语法因数据库而异。但上面给出了逻辑的想法。

在SQL Server中，语法为：

select distinct uid
from t
where exists (select 1
              from t t2
              where t2.uid = t.uid and
                    t2.date > t.date and
                    t2.date <= dateadd(second, 1, t.date)
             );

编辑：

或者，在SQL Server 2012+中，更快的替代方法是使用lead()或lag()：

select distinct uid
from (select t.*, lead(date) over (partition by uid order by date) as next_date
      from t
     ) t
where next_date < dateadd(second, 1, date);

如果你想要记录，而不仅仅是uid，那么你需要同时获得两者：

select t.*
from (select t.*,
             lag(date) over (partition by uid order by date) as prev_date,
             lead(date) over (partition by uid order by date) as next_date
      from t
     ) t
where next_date <= dateadd(second, 1, date) or
      prev_date >= dateadd(second, -1, date);

SQL Server-查找特定日期内的所有记录（不是那么简单！）

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