用于返回每个ID的最新记录的SQL查询

Question

我有一个具有以下结构的表。

LocId Value1 Value1Date     
............................................. 
1  50  2012-10-20 14:21:00.000  
1  70  2012-10-21 14:21:00.000  
1  90  2012-10-22 14:21:00.000  
1  100  2012-10-23 14:21:00.000  
2  20  2012-10-20 14:21:00.000  
2  40  2012-10-21 11:21:00.000  
2  70  2012-10-22 14:21:00.000  
2  80  2012-10-23 14:21:00.000 
3  50  2012-10-20 14:21:00.000  
3  70  2012-10-21 11:21:00.000  
3  80  2012-10-22 14:21:00.000  
3  90  2012-10-23 14:21:00.000 

我想达到的目标是，为每个[LOCID]，我需要最新的日期时间的[值1]（即2012年10月23日）。返回的表格应如下所示：

LocId Value1 Value1Date     
............................................. 
1  100  2012-10-23 14:21:00.000 
2  80  2012-10-23 14:21:00.000 
3  90  2012-10-23 14:21:00.000 

请问谁能举手？由于

Answer 1

您可以使用排名在分区像这样：

select * from 
(select locid, value1, value1date, 
    rank() over (partition by locid order by value1date desc) as rank 
    from table1) t 
where t.rank=1 

查看关于让所有的ID和循环通过他们是非常低效的SqlFiddle

Answer 2

这可以使用Common Table Expression实现和Window Function

WITH records 
AS 
(
    SELECT LocID, VAlue1, Value1Date, 
      ROW_NUMBER() OVER (PARTITION BY LocID ORDER BY Value1Date DESC) rn 
    FROM TableName 
) 
SELECT LocID, VAlue1, Value1Date 
FROM records 
WHERE rn = 1 

• SQLFiddle Demo
• TSQL Ranking Functions

Answer 3

这种方法会奏效。

select locid, value1, value1date 
from yourtable join 
(select locid id, max(value1date) maxdate 
from yourtable 
group by locid) temp on id = locid 
and value1date = maxdate 

Answer 4

这是执行你要找的东西的mysql查询。我不太了解其他SQL。

SELECT * FROM table WHERE LocID = n ORDER BY Value1Date DESC LIMIT 1; 

如果你从一个程序查询，你可以使用：

SELECT LocID FROM table ORDER BY LocID DESC LIMIT 1; 

获得ID的总数（假设他们是连续的），然后把它放在一个循环获取所有最近的日期。

希望有所帮助。