sql - 获取最新的不同记录

Question

考虑下表：

用户创建日期时间数量
----- ----- --------
吉姆 2012-09-19 01:00 1
吉姆 2012-09-19 02:00 5
吉姆 2012-09-19 03:00 2
鲍勃 2012-09-19 02:00 2
鲍勃 2012-09-19 03:00 9
鲍勃 2012-09-19 05:00 1

什么查询将返回每个 的最新行（由 定义CreatedDateTime）User，以便我们可以确定关联的Quantity.

即以下记录

用户创建日期时间数量
----- ----- --------
吉姆 2012-09-19 03:00 2
鲍勃 2012-09-19 05:00 1

我们认为我们可以简单地Group By User添加CreatedDateTime一个Having MessageCreationDateTime = MAX(.MessageCreationDateTime. 当然这不起作用，因为Quantity在Group By.

Answer 1

由于您使用的是 SQL Server，因此您可以在此使用窗口函数。

SELECT [User], CreatedDateTime, Quantity
FROM
    (
        SELECT [User], CreatedDateTime, Quantity,
                ROW_NUMBER() OVER(PARTITION BY [User] ORDER BY CreatedDateTime DESC) as RowNum
        FROM tableName
    ) a
WHERE  a.RowNum = 1

SQLFiddle 演示

Answer 2

;WITH x AS
(
  SELECT [User], CreatedDateTime, Quantity, 
   rn = ROW_NUMBER() OVER (PARTITION BY [User] ORDER BY CreatedDateTime DESC)
  FROM dbo.table_name
)
SELECT [User], CreatedDateTime, Quantity
FROM x WHERE rn = 1;

Answer 3

如果您没有使用窗口函数的能力，那么您可以使用子查询：

select t1.[user], t2.mxdate, t1.quantity
from yourtable t1
inner join
(
    select [user], max(CreatedDateTime) mxdate
    from yourtable
    group by [user]
) t2
    on t1.[user]= t2.[user]
    and t1.CreatedDateTime = t2.mxdate

见SQL Fiddle with Demo

Answer 4

SELECT DISTINCT 
User, 
CreatedDateTime, 
Quantity
FROM 
YourTable
WHERE 
CreatedDateTime = 
(SELECT MAX(CreatedDateTime) FROM YourTable t WHERE t.User = YourTable.User)

Answer 5

select * from <table_name> where CreatedDateTime in (select max(CreatedDateTime) from <table_name> group by user) group by user;