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

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：

Cannot perform an aggregate function on an expression containing an aggregate or a subquery.

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

想象一下,这个查询的结果将被交给一个想知道一天中最忙碌时间是什么的PHB.

> SQL Server 2005

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)