用於返回每個ID的最新記錄的SQL查詢

Question

我有一個具有以下結構的表。

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 

我想達到的目標是，爲每個[LOCID]，我需要最新的日期時間的[值1]（即2012年10月23日）。返回的表格應如下所示：

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 

請問誰能舉手？由於

Answer 1

您可以使用排名在分區像這樣：

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

查看關於讓所有的ID和循環通過他們是非常低效的SqlFiddle

Answer 2

這可以使用Common Table Expression實現和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

這種方法會奏效。

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

這是執行你要找的東西的mysql查詢。我不太瞭解其他SQL。

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

如果你從一個程序查詢，你可以使用：

SELECT LocID FROM table ORDER BY LocID DESC LIMIT 1; 

獲得ID的總數（假設他們是連續的），然後把它放在一個循環獲取所有最近的日期。

希望有所幫助。