Question

我的SQL Server表如下所示

 Id     Date                Msg
 1    01-01-2015             A
 1    02-01-2015             B
 2    03-01-2015             C

我想写一个查询来获得结果如下

   Id          Date            Msg
    1           02-01-2015      B
    2           03-01-2015      C

我正在尝试

 SELECT Id,Max(Date) , Msg
 Group By Id, Msg

这也是我的第一行。我希望Msg for Max date为特定ID。

任何建议请

Answer 1

使用Window Function

SELECT Id,
       Date,
       Msg
FROM   (SELECT *,
               Row_number()OVER (partition BY Id ORDER BY [Date] DESC ) rn
        FROM   tablename) a
WHERE  rn = 1

另一种方法是在date中找到每Id个subquery，然后在主表中找到join结果。试试这个。

SELECT Id,
       Date,
       Msg
FROM   tablename a
       JOIN (SELECT Id,
                    Max([Date])[Date]
             FROM   tablename
             GROUP  BY Id) B
         ON a.Id = b.id
            AND a.[Date] = b.[Date]

Answer 2

在where子句

中获取最大日期行的简单方法

这是简单的示例查询;

SELECT * FROM   TABLE  T1  
WHERE  T1.TIME = (SELECT MAX(TIME) FROM TABLE  WHERE ID = T1.ID )

如何在Group By中获取不同的记录

2 个答案: