Question

说我有一张桌子

表格播放

date     | track_id  | user_id | rating
-----------------------------------------
20170416 | 1         | 1       | 3  (***)
20170417 | 1         | 1       | 5
20170418 | 2         | 1       | 1
20170419 | 3         | 1       | 4
20170419 | 3         | 1       | 2  (***)
20170420 | 1         | 2       | 5

我想要做的是每个唯一的track_id，user_id我想要最高评级行。即在下面的表中生成（***）行被删除。

20170417 | 1         | 1       | 5
20170418 | 2         | 1       | 1
20170419 | 3         | 1       | 2
20170420 | 1         | 2       | 5

知道一个合理的SQL查询是做什么的吗？

Answer 1

使用MAX内置函数和GROUP by子句：

    SELECT track_id, user_id, MAX(rating)
    FROM Your_table
    GROUP BY track_id, user_id;

Answer 2

假设您也想要max日期，请尝试以下操作：

SELECT track_id, user_id, MAX(rating), MAX(date)
FROM plays
GROUP BY track_id, user_id;

Answer 3

select * from [table] t1
inner join
(
select track_id, user_id, max(rating) maxRating
from [table]
group by track_id, user_id
) tmp
on t1.track_id = tmp.track_id
and t1.user_id = tmp.user_id
and t1.rating = tmp.maxRating;

Answer 4

如果您有GROUP BY子句中未提及的额外列，则可以使用以下子查询：

SELECT track_id,user_id,rating  --, other columns you want to display
FROM(
      SELECT track_id, user_id, rating,ROWNUMBER() OVER(PARTITION BY track_id, user_id ORDER BY rating DESC) as RN --,Other columns which you want to display
       FROM Your_table 
                        ) X
WHERE X.RN=1

SQL从每个组中获取最大值

4 个答案: