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mysql - 根据 SUM 为行分配位置值

问题描述

我正在使用下面的代码来获取人们注册并输入数据的团队的总距离，然后为团队分配一个位置。

SELECT @curRow := @curRow + 1 AS position, ROUND(SUM(d.dist_activity_duration 
             * CASE 
                 WHEN d.dist_is_distance = 0 THEN s.activity_steps / 2000 
                 WHEN d.dist_is_distance = 1 THEN 1 
               END)
              ,2)   AS miles, t.team_name AS team_name
                    FROM distance d     
                    JOIN    (SELECT @curRow := 0) r 
                    JOIN activities a 
                    ON a.id = d.dist_activity_id
                    JOIN steps s
                    ON s.id = a.steps_id
                    JOIN members AS m   
                    ON d.member_id = m.id
                    JOIN teams AS t 
                    ON t.id = m.member_team_id
                    GROUP BY team_name 
                    ORDER BY miles DESC

上面的代码输出以下结果

position    miles    team_name
2           134.05   team 1
1           78.00    team 2

我希望将位置 1 分配给里程数最高的团队，将位置 2 分配给第二高的团队……以此类推。

标签： mysqlsql

解决方案

在 MySQL 8+ 中，您只需使用row_number()：

SELECT ROW_NUMBER() OVER (ORDER BY miles DESC) AS position, t.*
FROM (SELECT ROUND(SUM(d.dist_activity_duration *
                       CASE WHEN d.dist_is_distance = 0 THEN s.activity_steps / 2000 
                            WHEN d.dist_is_distance = 1 THEN 1 
                       END), 2)  AS miles, t.team_name AS team_name
      FROM distance d JOIN   
           activities a 
           ON a.id = d.dist_activity_id JOIN
           steps s
           ON s.id = a.steps_id JOIN
           members m   
           ON d.member_id = m.id JOIN
           teams t 
           ON t.id = m.member_team_id
      GROUP BY team_name 
     ) t
ORDER BY miles DESC;

早期版本的 MySQL 支持变量，但它们不能很好地与GROUP BY和一起使用ORDER BY。解决方案是子查询（如上）：

SELECT (@rn := @rn + 1) AS position, 
FROM (SELECT ROUND(SUM(d.dist_activity_duration *
                       CASE WHEN d.dist_is_distance = 0 THEN s.activity_steps / 2000 
                            WHEN d.dist_is_distance = 1 THEN 1 
                       END), 2)  AS miles, t.team_name AS team_name
      FROM distance d JOIN   
           activities a 
           ON a.id = d.dist_activity_id JOIN
           steps s
           ON s.id = a.steps_id JOIN
           members m   
           ON d.member_id = m.id JOIN
           teams t 
           ON t.id = m.member_team_id
      GROUP BY team_name 
      ORDER BY miles DESC
     ) t CROSS JOIN
     (SELECT @rn := 0) params;

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