Question

我有这个查询

SELECT DISTINCT u.fbid, u.name,r.points
        FROM users u, players_records r
            WHERE u.fbid = r.user_id
ORDER BY r.points DESC LIMIT 5

我想获得顶级玩家但只有不同的5名玩家，此查询无法正常显示重复的用户ID 有什么帮助吗？

以上查询的结果

1112222 Name 1 9310
3334444 Name 2 8380
3334444 Name 2 7010
5555666 Name 3 6080
1112222 Name 1 4890

所以ids重复

Answer 1

您可以选择最高分数并按用户分组。

SELECT u.fbid, u.name, MAX(r.points) max_points
FROM users u, player_records r
WHERE u.fbid = r.user_id
GROUP BY u.fbid, u.name
ORDER BY max_points DESC LIMIT 5

Answer 2

听起来你想要每个用户的最大点数。所以你可以这样做：

SELECT
    users.fbid, 
    users.name,
    maxRecords.points
FROM
    users
    JOIN
    (
        SELECT
            MAX(players_records.points) AS points,
            players_records.user_id
        FROM
            players_records
        GROUP BY
            players_records.user_id
    ) AS maxRecords
    ON maxRecords.user_id=users.fbid
ORDER BY 
    maxRecords.points DESC 
LIMIT 5

如果我了解你的数据。然后输出将是这样的：

1112222 Name 1 9310
3334444 Name 2 8380
5555666 Name 3 6080

Answer 3

如果你需要前五名球员：

SELECT TOP 5 u.fbid, u.name, max(r.points) AS points
FROM users u
LEFT JOIN players_records r ON u.fbid = r.user_id
GROUP BY u.fbid, u.name
ORDER BY points DESC

如果你需要按SUM积分排名前5位球员：

SELECT TOP 5 u.fbid, u.name, SUM(r.points) AS points
FROM users u
LEFT JOIN players_records r ON u.fbid = r.user_id
GROUP BY u.fbid, u.name
ORDER BY points DESC

Answer 4

这样的事情可能与你想要的一样。

select fbid, u.name, sum(r.points) totalpoints
from users u join players_records r on u.fbid = r.user_id
group by fbid, u.name
order by totalpoints desc
limit 5

DISTINCT不在内连接中工作

4 个答案: