我有这个SQL：
SELECT (Min(Time_In) & " to " & Max(Time_Out)) AS [Week Of],
Round((Sum(DATEDIFF("n", Time_In, Time_Out))/Count(DATEDIFF("n", Time_In, Time_Out))),2) AS [Avg Min of Jog]
FROM SomeTable
WHERE len(Time_In) > 0 AND len(Time_Out) > 0
AND #01/01/2012# <= Time_In AND #12/31/2012# >= Time_In
GROUP BY DatePart('ww',Time_In);
选择每周慢跑的平均时间。我还要计算每周慢跑多少次，我想通过计算jog_id SomeTable jog_id可能有一个SELECT (Min(Time_In) & " to " & Max(Time_Out)) AS [Week Of],
Round((Sum(DATEDIFF("n", Time_In, Time_Out))/Count(DATEDIFF("n", Time_In, Time_Out))),2) AS [Avg Min of Jog],
Count(distinct jog_id) AS [Num Jogs]
FROM SomeTable
WHERE len(Time_In) > 0 AND len(Time_Out) > 0
AND #01/01/2012# <= Time_In AND #12/31/2012# >= Time_In
GROUP BY DatePart('ww',Time_In);
的5个条目来计算。
我试过了：
Syntax error (missing operator) in query expression 'Count(distinct jog_id)'.但是这给了我{{1}}
我错过了什么？
答案 0 :(得分：3)
如果不使用子查询，则无法在Access中执行此操作 EX：
SELECT Count(*)
FROM
(SELECT DISTINCT Name FROM table1);
Here是关于该主题的详细文章：
答案 1 :(得分：-1)
Access几乎在所有情况下都拒绝行事，其中一个是Count(distinct x)的支持。