TSql Sum By Date

Question

我试过看一些例子，但没有一个与我想做的事情有关。在我的表中，我有一个具有日期的列，我想要做的是按日期添加记录。例如，在3月12日，13日，14日等添加所有记录。因此每天应该有它的日期和计算的总和。我该如何应用？

SELECT DTTM
FROM 
[Audits].[dbo].[Miscount]
Group by DTTM
Order by DTTM desc

干杯！

Answer 1

如果您想要每天记录的数量：

SELECT DTTM,COUNT(*) AS Total
FROM 
[Audits].[dbo].[Miscount]
Group by DTTM
Order by DTTM desc

或者，如果您想要每条记录上的字段总和：

SELECT DTTM,SUM(field1) AS Sum
FROM 
[Audits].[dbo].[Miscount]
Group by DTTM
Order by DTTM desc

或者，如果DTTM是日期时间，那么您可以使用：

SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, DTTM)) AS DTTM,COUNT(*) AS Total
FROM 
[Audits].[dbo].[Miscount]
Group by DATEADD(dd, 0, DATEDIFF(dd, 0, DTTM))
Order by DATEADD(dd, 0, DATEDIFF(dd, 0, DTTM)) desc

较新版本的SQL Sever将支持Date类型，因此您可以这样做：

SELECT CAST(DTTM AS Date) AS DTTM,COUNT(*) AS Total
FROM 
[Audits].[dbo].[Miscount]
Group by CAST(DTTM AS Date)
Order by CAST(DTTM AS Date) desc

Answer 2

如果DTTM是DateTime类型，那么

SELECT
            CAST([DTTM] AS Date) [Date],
            COUNT(*) [Total]
    FROM 
            [Audits].[dbo].[Miscount]
    GROUP BY
            CAST([DTTM] AS Date)
    ORDER BY
            1 DESC

会这样做。 Heres some fiddle来演示。