如何在SQL中对我的列进行分组？

Question

如何按时间分组？我想这一点，但它给错误“无效列名‘时间’。”：

select Count(Page) as VisitingCount, CONVERT(VARCHAR(5), Date, 108) as [Time] 
from scr_SecuristLog 
where Date between '2009-05-04 00:00:00' and '2009-05-06 14:58' 
and [user] in (select USERNAME from scr_CustomerAuthorities) 
group by [Time] order by [VisitingCount] asc 

Answer 1

[时间]是一个列别名。尝试

SELECT 
     COUNT(Page) AS VisitingCount 
    , CONVERT(VARCHAR(5),Date, 108) AS [Time] 
FROM 
    scr_SecuristLog 
WHERE 
    Date BETWEEN '2009-05-04 00:00:00' AND '2009-05-06 14:58' 
    AND 
    [user] IN (
       SELECT 
        USERNAME    
       FROM 
        scr_CustomerAuthorities 
       ) 
GROUP BY 
    CONVERT(VARCHAR(5),Date, 108) 
ORDER BY 
    [VisitingCount] ASC 

Answer 2

尝试

GROUP BY CONVERT(VARCHAR(5),Date, 108) 

始终在您的SELECT子句确保你按的一切，没有一个它的聚合函数。

Answer 3

select Count(Page) as VisitingCount,CONVERT(VARCHAR(5),Date, 108) as [Time] from scr_SecuristLog 
where Date between '2009-05-04 00:00:00' and '2009-05-06 14:58' 
and [user] in(select USERNAME    
    from scr_CustomerAuthorities) 
group by CONVERT(VARCHAR(5),Date, 108) order by [VisitingCount] asc 

我修改GROUP BY以包括用于[Time]实际表达，而不是列的别名（因为这不能在GROUP BY可以使用，只有ORDER BY）

Answer 4

看起来像日期是这里的一列，但它是一个关键字，没有引用。也许这是个问题（不是虽然测试）：

select Count(Page) as VisitingCount,CONVERT(VARCHAR(5),[Date], 108) as [Time] from scr_SecuristLog 
where [Date] between '2009-05-04 00:00:00' and '2009-05-06 14:58' 
and [user] in(select USERNAME    
    from scr_CustomerAuthorities) 
group by [Time] order by [VisitingCount] asc 

Answer 5

如果你不想重复日期转换（有时计算是有点比简单的转换更密集），那么你可以使用这样的事情：

select * 
from (select Count(page) .., ... As [Date] from ... where ...) UG 
group by UG.[Date] 

请注意，您有带来的“UG”名内部选择，并认为这将是最有可能的，而且在大多数情况下比原分组重复转换表达效率较低。 你的整个表情很可能会改变......但是，只是你知道这是可能的。