Question

我正在使用下面的代码来获取人们已经注册并输入数据的团队的总距离，然后为该团队分配职位。

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分配给里程最高的团队，将位置2分配给第二高的团队...等等。

Answer 1

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;

MySQL的早期版本支持变量，但是它们不能与GROUP BY和ORDER BY一起使用。解决方案是一个子查询（如上所述）：

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;

Answer 2

这对我有用。

SELECT (@rn := @rn + 1) AS position, team_name, miles
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 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
 ) t CROSS JOIN
 (SELECT @rn := 0) params;

根据SUM将位置值分配给行

2 个答案: