TSQL:无法在COUNT(*)上执行聚合函数AVG以查找一天中最繁忙的时段

Question

考虑一个包含日志数据的SQL Server表.重要的部分是:

CREATE TABLE [dbo].[CustomerLog](
    [ID] [int] IDENTITY(1,1) NOT NULL,
    [CustID] [int] NOT NULL,
    [VisitDate] [datetime] NOT NULL,
 CONSTRAINT [PK_CustomerLog] PRIMARY KEY CLUSTERED ([ID] ASC)) ON [PRIMARY]

这里的查询是围绕当天查找每小时的访问分布.我们有兴趣查看给定日期范围内每小时平均访问次数的分布情况.

查询结果将是这样的:

HourOfDay   Avg.Visits.In.Hour
0            24
1            16
5            32
6            89
7           823
etc.etc.

目的是写一个这样的查询:

SELECT  DATEPART(hh, VisitDate)
        ,AVG(COUNT(*))    
FROM    CustomerLog
WHERE   VisitDate   BETWEEN 'Jan 1 2009' AND 'Aug 1 2009'
GROUP BY   DATEPART(hh, VisitDate)

但这不是有效的查询:

无法对包含聚合或子查询的表达式执行聚合函数.

问题:如何重新编写此查询以收集平均总数(即代替AVG(COUNT(*))小时？

想象一下,这个查询的结果会传递给想要知道一天中最忙碌时间的PHB.

• SQL Server 2005+

Answer 1

使用内联视图:

SELECT DATEPART(hh, x.visitdate),
       AVG(x.num)
  FROM (SELECT t.visitdate,
               COUNT(*) 'num'
          FROM CUSTOMERLOG t
         WHERE t.visitdate BETWEEN 'Jan 1 2009' AND 'Aug 1 2009'
      GROUP BY t.visitdate) x
GROUP BY DATEPART(hh, x.visitdate)

使用CTE(SQL Server 2005+)等效:

WITH visits AS (
   SELECT t.visitdate,
          COUNT(*) 'num'
     FROM CUSTOMERLOG t
    WHERE t.visitdate BETWEEN 'Jan 1 2009' AND 'Aug 1 2009'
 GROUP BY t.visitdate)
   SELECT DATEPART(hh, x.visitdate),
         AVG(x.num)
    FROM visits x
GROUP BY DATEPART(hh, x.visitdate)