按星期几每小时平均计数

Question

这个question帮助我进入了那里：

SELECT 
    [Day], 
    [Hour], 
    [DayN], 
    AVG(Totals) AS [Avg]
FROM
(
    SELECT 
        w = DATEDIFF(WEEK, 0, ForDateTime),
        [Day]  = DATENAME(WEEKDAY, ForDateTime),
        [DayN] = DATEPART(WEEKDAY, ForDateTime),
        [Hour] = DATEPART(HOUR,    ForDateTime),
        Totals = COUNT(*)
    FROM 
        #Visit
    GROUP BY 
        DATEDIFF(WEEK, 0, ForDateTime),
        DATENAME(WEEKDAY, ForDateTime),
        DATEPART(WEEKDAY, ForDateTime),
        DATEPART(HOUR,    ForDateTime)
) AS q
GROUP BY 
    [Day], 
    [Hour], 
    [DayN]
ORDER BY 
    DayN; 

如何更改此值，而不是按小时显示平均值，例如9、10、11、12等。它一直显示到09：30-10：30,10：30-11：30,11：30-12：30,12：30-13：30一直到23:30。

Answer 1

一种简单的方法是将ForDateTime偏移30分钟。基本上，您只需要在查询中将ForDateTime的每次出现替换为dateadd(minute, 30, ForDateTime)。

在结果集中，Hour 9为您提供了从8:30到9:30的时隙，依此类推。