Question

这是我的表和查询

表1

AsOfDate                  Value
2015-06-06 11:08:21.247     5 
2015-06-06 12:08:21.247     7 
2015-06-07 10:08:21.247     8 
2015-06-08 11:08:21.247     8      
2015-06-08 10:09:21.247     9 

SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate])) as DateOnly ,Min([Value])
FROM Table1 group by DateOnly

我收到错误消息“无效的列名'DateOnly'。”

实现它的最佳方法是什么？

Answer 1

更改您的查询：

SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate])) as DateOnly ,Min([Value])
FROM Table1 group by DateOnly

要

SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate])) as DateOnly ,Min([Value])
FROM Table1 group by DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate]))

别名DateOnly已知为查询级别（仅限于其顶部）

Answer 2

你必须group by DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate]))。很遗憾，您无法在group by子句中使用别名。

Answer 3

此处DateOnly是列别名。因此，请在GROUP BY中使用表达式本身

SELECT DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate])) as DateOnly ,Min([Value])
FROM Table1 
group by DATEADD(dd, 0, DATEDIFF(dd, 0, [AsOfDate]))

Answer 4

您没有使用从datetime数据中删除时间部分的最佳方法。如果您使用支持date数据类型的SQL Server版本，只需使用CAST(AsOfDate as date)，或者您可以使用：CAST(CAST(AsOfDate as int) as datetime)

所以查询可以是这样的：

SELECT CAST(AsOfDate as date) as DateOnly ,Min([Value])
FROM Table1
GROUP BY CAST(AsOfDate as date)

分组依据和DateAdd以及简单的计算SQL Server

4 个答案: