SQL 多行排名 [相同的用户仍将具有相同的排名]

Question

我对 SQL 了解一点，我一直想知道如何从我的 Users table 中 select 用户，我发现了这个：

SELECT *, FIND_IN_SET( Score, (    
    SELECT GROUP_CONCAT( Score ORDER BY Score DESC ) 
        FROM Users
    )
) AS Rank
FROM Users
WHERE UserID = 100
ORDER BY Score DESC

问题是，如果同一用户有多个行，那么他们将单独排名。我想知道如何获得最高的 Score 并拥有 Rank return 那

它看起来像什么（没有 WHERE UserID = 100 子句）：

ID-----UserID-----Score----Rank
1------100--------6--------1
2------192--------4--------2
3------192--------3--------3

我想要的（没有 WHERE UserID = 100 子句）：

ID-----UserID-----Score----Rank
1------100--------6--------1
2------192--------4--------2
3------192--------3--------2

感谢您的帮助！

Answer 1

尝试以下操作：

SET @rank=0;
SELECT *, @rank:=@rank+1 AS rank
FROM Users
WHERE UserID = 100
ORDER BY Score DESC

Answer 2

使用 GROUP BY(UserID) 和 MAX(Score) 得到 maxScore，然后用它排名。

（作为 OP 希望每一行用户，但只有最好的排名，我们添加一个 INNER JOIN）

SELECT TM.`UserID`, TM.`Score`, T3.`rank`
FROM `Users` as TM
INNER JOIN (
        SELECT `UserID`, FIND_IN_SET(
                Max(`Score`), (
                    select GROUP_CONCAT(
                        DISTINCT `Score`
                        order by `Score` DESC
                    )
                    From (
                        select MAX(`Score`) as `Score`
                        FROM `Users` as T0
                        GROUP BY `UserID`
                    ) as T1
                )
            ) as `rank`
        FROM `Users` as T2
        group by `UserID`
    ) as T3
    ON TM.`UserID` = T3.`UserID`
-- If we need to filter or sort, we make it here :
WHERE TM.`UserID` = 192
ORDER BY TM.`Score`

Answer 3

您可以统计每个得分较低的用户：

SELECT u.ID, u.UserID u.Score, 1 + COUNT(*) as Rank -- 1 + count every user with less score
FROM Users u
JOIN Users loser
  ON loser.Score < u.score -- join with every user having less score
GROUP BY u.ID, u.UserID, u.Score
ORDER BY Rank

SQL 多行排名 [相同的用户仍将具有相同的排名]

SQL Ranking with multiple rows [Same users will still have same rank ]

mysql

rank