滞后函数和 SUM答案

【问题标题】：Lag functions and SUM滞后函数和 SUM
【发布时间】：2020-11-04 00:16:24
【问题描述】：

我需要获取每天至少离线 20 分钟的用户列表。这是我的数据

我有这个起始查询，但我被困在如何对offline_mins 的差异求和，即需要在 where 子句中添加“and sum(offline_mins)>=20”

SELECT  
   userid, 
    connected,
   LAG(recordeddt) OVER(PARTITION BY userid
   ORDER BY userid, 
            recordeddt) AS offline_period,
            DATEDIFF(minute, LAG(recordeddt) OVER(PARTITION BY userid
   ORDER BY userid, 
            recordeddt),recordeddt)  offline_mins
FROM device_data where connected=0;

我的预期结果：

提前致谢。

【问题讨论】：

请出示您的预期结果。
(1) 每个用户每 5 分钟是否总是有一条记录？ (2) 在“每天”下线至少20分钟的用户中定义“每天”。
每天=24 小时。可能不是每 5 分钟记录一次，但是当用户最终上线时，会有一条 connected=1 的记录

标签： sql sql-server datetime gaps-and-islands date-arithmetic

【解决方案1】：

这看起来像是一个间隙和孤岛问题，您希望将具有相同用户 ID 和状态的相邻行组合在一起。

首先，这是一个计算岛屿的查询：

select userid, connected, min(recordeddt) startdt, max(lead_recordeddt) enddt,
    datediff(min(recordeddt), max(lead_recordeddt)) duration
from (
    select dd.*,
        row_number()     over(partition by userid order by recordeddt) rn1,
        row_number()     over(partition by userid, connected order by recordeddt) rn2,
        lead(recordeddt) over(partition by userid order by recordeddt) lead_recordeddt
    from device_data dd
) dd
group by userid, connected, rn1 - rn2

现在，假设您希望用户每天离线至少 20 分钟。您可以每天细分岛屿，并使用having 子句进行过滤：

select userid
from (
    select recordedday, userid, connected,
        datediff(min(recordeddt), max(lead_recordeddt)) duration
    from (
        select dd.*, v.*,
            row_number()     over(partition by v.recordedday, userid order by recordeddt) rn1,
            row_number()     over(partition by v.recordedday, userid, connected order by recordeddt) rn2,
            lead(recordeddt) over(partition by v.recordedday, userid order by recordeddt) lead_recordeddt
        from device_data dd
        cross apply (values (convert(date, recordeddt))) v(recordedday)
    ) dd
    group by convert(date, recordeddt), userid, connected, rn1 - rn2
) dd
group by userid
having count(distinct case when connected = 0 and duration >= 20 then recordedday end) = count(distinct recordedday)

【讨论】：

【解决方案2】：

如前所述，这是一个差距和孤岛问题。这是我对它的看法，它使用一个简单的滞后函数来创建组，过滤掉连接的行，然后处理日期范围。

CREATE TABLE #tmp(ID int, UserID int, dt datetime, connected int)
INSERT INTO #tmp VALUES
(1,1,'11/2/20 10:00:00',1),
(2,1,'11/2/20 10:05:00',0),
(3,1,'11/2/20 10:10:00',0),
(4,1,'11/2/20 10:15:00',0),
(5,1,'11/2/20 10:20:00',0),
(6,2,'11/2/20 10:00:00',1),
(7,2,'11/2/20 10:05:00',1),
(8,2,'11/2/20 10:10:00',0),
(9,2,'11/2/20 10:15:00',0),
(10,2,'11/2/20 10:20:00',0),
(11,2,'11/2/20 10:25:00',0),
(12,2,'11/2/20 10:30:00',0)


SELECT UserID, connected,DATEDIFF(minute,MIN(DT), MAX(DT)) OFFLINE_MINUTES 
FROM
(
    SELECT *, SUM(CASE WHEN connected <> LG THEN 1 ELSE 0 END) OVER (ORDER BY UserID,dt) grp
    FROM
    (
        select *, LAG(connected,1,connected) OVER(PARTITION BY UserID ORDER BY UserID,dt) LG
        from #tmp
    ) x
) y
WHERE connected <> 1
GROUP BY UserID,grp,connected
HAVING DATEDIFF(minute,MIN(DT), MAX(DT)) >= 20

【讨论】：