Mysql 从地图和计数中获得最佳效果

我给你看一个我需要的例子

包含以下列和数据的表，例如：

id  | mapname   | username  | score
----|-----------|-----------|--------
1   | mid_town  | carolina  | 5
2   | mid_town  | rupert    | 10
3   | id_room   | carolina  | 7
4   | mid_town  | mordor    | 14
5   | id_room   | intel     | 12
6   | id_room   | puyol     | 21
7   | pindle    | puyol     | 17
8   | pindle    | messi     | 28
9   | mid_town  | puyol     | 24
n   | unknown   | unknown   | unknown

我需要最好的从地图名称顺序分数描述，但我不知道所有的地图名称

那么在这种情况下,

mid_镇最好的是普约尔，24岁

id_房间最好的是普约尔，21岁

平德尔最好的球员是梅西，28岁

输出：

puyol 2
messi 1

如何在一个查询中转换它？

尝试以下方法：

select username, score from `table` where (mapname,score) in (select mapname, max(score) from `table` group by mapname)

试试这个，伙计：

SELECT
    username, score
FROM
    <your_table>
WHERE
    mapname = 'midtown'
HAVING
    score = MAX(SCORE)

UNION

SELECT
    username, score
FROM
    <your_table>
WHERE
    mapname = 'id_room'
HAVING
    score = MAX(SCORE)

UNION

SELECT
    username, score
FROM
    <your_table>
WHERE
    mapname = 'pindle'
HAVING
    score = MAX(SCORE);  

或者这个：

SELECT
    username, score
FROM
    <your_table>
GROUP BY
    mapname
ORDER BY
    score DESC;

一位朋友回答说：

SELECT 
     t1.username, COUNT(t3.var2) as var1
FROM 
     data AS t1
JOIN
    (SELECT 
        t2.mapname, MAX(t2.score) as var2
        FROM data AS t2
        GROUP BY mapname) AS t3
ON t1.mapname = t3.mapname
AND t1.score = t3.var2
GROUP BY t1.username
ORDER BY var1 DESC;

---

我从五分钟前开始执行查询，我还在等待，有什么问题吗？[…]但我不知道所有的地图名称。试试另一个，伙计。