このクエリはどのように評価できますか？

Question

このクエリでエラーが発生します。メッセージ156、レベル15、状態1、行10 キーワード 'where'の近くに構文が正しくありません。

declare @date1 datetime,@date2 datetime , @COUNT INT , @countgap int, @order int 
seLECT @date1='2009-05-11' , @date2 = '2009-05-12' 
seLECT @countgap = 30 , @COUNT = 0, @order=23 

select VisitingCount from (
select count(page) as VisitingCount, 
    (datepart(hour,Date)*60+datepart(minute,Date))/@countgap as OrderNumber 
    from scr_SecuristLog 
    where Date between @date1 and @date2 
    GROUP BY (datepart(hour,Date)*60+datepart(minute,Date))/@countgap) where OrderNumber=@order

Answer 1

Pherhaps：

GROUP BY (datepart(hour,Date)*60+datepart(minute,Date))/@countgap) 
HAVING OrderNumber=@order 

か：あなたは、例えば、あなたの派生テーブルの名前を与える必要が

where (Date between @date1 and @date2) AND OrderNumber=@order 
GROUP BY (datepart(hour,Date)*60+datepart(minute,Date))/@countgap) 

Answer 2

DT1。ここでは、私の個人的な好みにSQLテキストを再フォーマットしました:)読みやすくするために：

select DT1.VisitingCount 
    from (
     select count(page) as VisitingCount, 
       (datepart(hour,Date)*60+datepart(minute,Date))/@countgap 
        as OrderNumber 
      from scr_SecuristLog 
     where Date between @date1 and @date2 
     GROUP 
      BY (datepart(hour,Date)*60+datepart(minute,Date))/@countgap 
     ) AS DT1 
where DT1.OrderNumber=@order