Question

我有一个查询，我在其中确定用户提交的特定表单超过1次：

select userid, form_id, count(*)
from table_A
group by userid, form_id
having count(userid) > 1

但是，我试图查看哪些用户在5秒的时间范围内提交了多个表单（我们在此表中有一个提交时间戳的字段）。如何根据该标准缩小此查询范围？

Answer 1

一种方法是通过DATEDIFF(Second, '2017-01-01', SubmittionTimeStamp) / 5添加到组中这将根据userid，form_id和五秒间隔对记录进行分组：

select userid, form_id, count(*)
from table_A
group by userid, form_id, datediff(Second, '2017-01-01', SubmittionTimeStamp) / 5
having count(userid) > 1

请阅读this SO post以获取更详细的说明。

Answer 2

@nikotromus

您尚未提供有关您的架构和其他可用列的大量详细信息，也未提供有关此信息的使用方式/方式和位置的详细信息。

但是，如果你想这样做，那就是＆＃34;生活＆＃34;所以将您的时间结果与当前时间戳进行比较，它看起来像：

SELECT userid, form_id, count(*)
  FROM table_A
 WHERE DATEDIFF(SECOND,YourColumnWithSubmissionTimestamp, getdate()) <= 5
GROUP BY userid, form_id
HAVING count(userid) > 1

Answer 3

您可以使用lag形成彼此相距5秒的行组，然后对它们进行聚合：

select distinct userid,
    form_id
from (
    select t.*,
        sum(val) over (
            order by t.submission_timestamp
            ) as grp
    from (
        select t.*,
            case 
                when datediff(ms, lag(t.submission_timestamp, 1, t.submission_timestamp) over (
                            order by t.submission_timestamp
                            ), t.submission_timestamp) > 5000
                    then 1
                else 0
                end val
        from your_table t
        ) t
    ) t
group by userid,
    form_id,
    grp
having count(*) > 1;

有关更多解释，请参阅此答案：

Group records by consecutive dates when dates are not exactly consecutive

Answer 4

我只想使用exists来吸引用户：

select userid, form_id
from table_A a
where exists (select 1
              from table_A a2
              where a2.userid = a.userid and a2.timestamp >= a.timestamp and a2.timestamp < dateadd(second, 5, a.timestamp
            );

如果您想要点票，只需添加group by和count(*)。

如何在有限的时间范围内缩小计数查询？

4 个答案: