選擇通過最新的時間戳

Question

我有如下表：

 
+-------+--------+---------------------+ 
| score | user |  scrap_time  | 
+-------+--------+---------------------+ 
| 200 | Bob | 2015-09-08 11:46:17 | 
+-------+--------+---------------------+ 
| 210 | Alice | 2015-09-08 11:46:17 | 
+-------+--------+---------------------+ 
| 240 | Bob | 2015-09-08 11:48:00 | 
+-------+--------+---------------------+ 
| 260 | Alice | 2015-09-08 11:48:01 | 
+-------+--------+---------------------+ 

我想所有的分數每個用戶是有最新scrap_time而忽略了舊的。

例子：

 
+-------+--------+---------------------+ 
| score | user |  scrap_time  | 
+-------+--------+---------------------+ 
| 240 | Bob | 2015-09-08 11:48:00 | 
+-------+--------+---------------------+ 
| 260 | Alice | 2015-09-08 11:48:01 | 
+-------+--------+---------------------+ 

我一直在試圖想出一個查詢，像這樣的：

select * from scores where date(scrap_time) = (select max(scrap_time) from scores); 

但是，這並沒有給我我需要的結果。

Answer 1

首先，你需要找到MAX（scrap_time）每用戶，然後找到那些行

SELECT * FROM scores WHERE (user,scrap_time) IN (
SELECT user,MAX(scrap_time) scrap_time FROM scores 
GROUP BY user); 

Answer 2

您需要獲得MAX(scrap_time)每user，然後只需返回到您的表以獲得所需的結果集。像下面這樣的東西應該工作：

SELECT s.score, s.user, s.scrap_time 
FROM scores AS s 
INNER JOIN (SELECT user, max(scrap_time) AS maxTime 
      FROM scores 
      GROUP BY user) AS t 
ON s.user = t.user AND s.scrap_time = maxTime 

Demo here

Answer 3

使用max函數與GROUP BY這將提供你最近每次得分的時間

select score, user, max(scrap_time) 
from  scores 
group by score;