Study From Work(2011-3-1)
时间:2011-03-03 来源:xiaoXD
解决方法:
方案一:
SELECT convert(char(10),datecolumn,120), count(*) FROM [SCM].[dbo].[feedback_detail] group by convert(char(10),datecolumn,120) order by 1
char(10)(是将日期的前十位字符分组)
方案二:
select DATEADD(day, datediff(day, 0, datecolumn), 0) , COUNT(*) FROM [SCM].[dbo].[feedback_detail] group by DATEADD(day, datediff(day, 0, datecolumn), 0) “day”可以换成别的日期单位
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