sql - sql查询根据给定的时间范围选择组内的第一次出现

Question

可能重复：
如何按小时或按 10 分钟分组时间

随着 MSSQL 得到这个结果：

从以下查询：

SELECT [DATESTAMP],[ZONE],[ALARM] 
FROM [UNIT].[dbo].[LOG] 
WHERE [DATESTAMP] BETWEEN @StartDate AND @EndDate 
    AND [ALARM] ='ERROR' 
    AND [ZONE] IN ('BOX 11','BOX 12') 
ORDER BY DATESTAMP DESC

我将如何更改查询以仅获取行 2,5,9,13,17,19,20 ？

这意味着我只想要跨越大约 1 分钟的组中的第一行（最旧的）。在该组内，而不会出现比 15 分钟更频繁的错误，Freewheelin

Answer 1

结束了：

SELECT  MIN([DATESTAMP]) as FirstOccurance FROM ....
WHERE ....
....
GROUP BY DATEPART(minute, [DATESTAMP])
ORDER BY FirstOccurance DESC

-再次感谢大家

Answer 2

如果我理解你，你想要每个特定分钟的最早条目，对吗？如果是这样，试试这个：

SELECT
    DateStamp,
    ZONE,
    Alarm
FROM
    [UNIT].[dbo].[LOG]  
    JOIN 
    (   SELECT 
            DateMinute = DATEADD(second, -DATEPART(second, DateStamp), DateStamp),
            Earliest = MIN(DateStamp)
        FROM
            [UNIT].[dbo].[LOG]  
        GROUP BY
            DATEADD(second, -DATEPART(second, DateStamp), DateStamp)
    ) X ON [UNIT].[dbo].[LOG].DateStamp = X.Earliest