Question

为了保持简单，请说我有一个SQL Server 2008表，其中一列具有日期时间数据类型。我想要一个产生每分钟间隔行数的查询。例如，结果如下所示：

7/3/2011 14:00:00 | 1000
7/3/2011 14:01:00 | 1097
7/3/2011 14:02:00 |  569

第一行意味着1000行的日期时间值介于7/3/2011 13:59:00和7/3/2011 14:00:00之间。

第二行意味着1097行的日期时间值介于7/3/2011 14:00:00和7/3/2011 14:01:00之间。

第三行意味着569行的日期时间值介于7/3/2011 14:01:00和7/3/2011 14:02:00之间。

谢谢。

Answer 1

此：

;WITH CTE_ExampleData as (
    select stamp = '07/07/2011 14:00:01'
    union select stamp = '07/07/2011 14:00:02'
    union select stamp = '07/07/2011 14:00:03'
    union select stamp = '07/07/2011 14:01:01'
    union select stamp = '07/07/2011 14:01:02'
    union select stamp = '07/07/2011 14:01:03'
    union select stamp = '07/07/2011 14:01:04'
    union select stamp = '07/07/2011 14:02:02'
    union select stamp = '07/07/2011 14:02:03'
    union select stamp = '07/07/2011 14:02:04'
    union select stamp = '07/07/2011 14:02:05'
)
 select 
    stamp = dateadd(mi,datediff(mi,0,stamp) + 1,0),
    rows = count(1)
 from CTE_ExampleData
 group by dateadd(minute,datediff(mi,0,stamp)+1,0)

返回

stamp                       rows
2011-07-07 14:01:00.000     3
2011-07-07 14:02:00.000     4
2011-07-07 14:03:00.000     4

Answer 2

简单组：

 Select DateAdd(minute, DateDiff(minute, 0, [colName]), 0), Count(*)
 From [tableName]
 Group By DateAdd(minute, DateDiff(minute, 0, [colName]), 0)

如果您希望输出中某个范围内的每一分钟，无论该分钟内是否有任何数据，请使用公用表表达式（CTE）：

 Declare @startMinute smalldatetime Set @startMinute = '30 June 2011'
 Declare @endMinute smalldatetime Set @endMinute = '1 July 2011';
 With minuteList(aMinute) As 
 (Select @startMinute Union All
    Select dateadd(minute,1, aMinute)
    From minuteList
    Where aMinute < @endMinute)
 Select aMinute, Count(T.[colName])
 From minuteList ml Left Join [tableName] T
      On DateAdd(minute, DateDiff(minute, 0, T.[colName]), 0) = aMinute
 Group By aMinute
 Option (MaxRecursion 10000);

Answer 3

如果有人想让你这样说，你也可以得到平均每分钟平均值，告诉他们你的ETL平均需要多长时间才能获得X记录。

SELECT AVG(ABC.TOTAL_MINUTE) FROM (Select DateAdd(minute, DateDiff(minute, 0,createdon),0) AS [DAY_MINUTE], Count(*) AS [TOTAL_MINUTE] From contactbase Group By DateAdd(minute, DateDiff(minute, 0, createdon), 0)) ABC here

或者，如果你想深入了解，你可以完成百分比......把它放在一个过程中......你明白了。

DECLARE @TtlToProcess AS DECIMAL --- put arbitrary number of 2.5 million in as this is aproximate, could be replaces with true source count if known. SET @TtlToProcess = 2500000 SELECT DATEDIFF(MINUTE,MIN(CB.CreatedOn),max(CB.CreatedOn)) AS [Minutes Run], ROUND(100*(CAST(COUNT(*) AS DECIMAL)/CAST(@TtlToProcess AS DECIMAL)),1) AS [%Complete], ROUND(100-(100*(CAST(COUNT(*) AS DECIMAL)/CAST(@TtlToProcess AS DECIMAL))),1) AS [% Left], COUNT(*) AS [Rows Processed], @TtlToProcess-COUNT(*) AS [Rows Left], (SELECT AVG(ABC.TOTAL_MINUTE) FROM (SELECT DATEADD(MINUTE, DATEDIFF(MINUTE, 0, CreatedOn), 0) AS [DAY_MINUTE], COUNT(*) AS [TOTAL_MINUTE] FROM ContactBase GROUP BY DATEADD(MINUTE, DATEDIFF(MINUTE, 0, CreatedOn), 0)) AS ABC) AS [Avg. Rows per Minute], ROUND(((@TtlToProcess-COUNT(*))/(SELECT AVG(ABC.TOTAL_MINUTE) FROM (SELECT DATEADD(MINUTE, DATEDIFF(MINUTE, 0, CreatedOn), 0) AS [DAY_MINUTE], COUNT(*) AS [TOTAL_MINUTE] FROM ContactBase GROUP BY DATEADD(MINUTE,DATEDIFF(MINUTE, 0, CreatedOn), 0)) ABC))/CAST(60 AS DECIMAL),2) AS [Est. Hours Left],DATEADD(hh,((@TtlToProcess-COUNT(*))/(SELECT AVG(ABC.TOTAL_MINUTE) FROM (SELECT DATEADD(MINUTE, DATEDIFF(MINUTE, 0, CreatedOn), 0) AS [DAY_MINUTE], COUNT(*) AS [TOTAL_MINUTE] FROM ContactBase GROUP BY DATEADD(MINUTE, DATEDIFF(MINUTE, 0, CreatedOn), 0)) AS ABC))/CAST(60 AS DECIMAL),GETDATE()) AS [Est. Complete Date_Time] FROM dbo.ContactBase AS CB

Answer 4

Select DateAdd(minute, DateDiff(minute, 0, createdon), 0) AS [DAY_MINUTE], Count(*) AS [TOTAL_MINUTE] From contactbase Group By DateAdd(minute, DateDiff(minute, 0, createdon), 0) order by DateAdd(minute, DateDiff(minute, 0, createdon), 0) desc

需要查询 - 确定每分钟的行数

4 个答案: