Присвоить значение позиции строке на основе суммы

Question

Я использую приведенный ниже код, чтобы получить общее расстояние для команд, в которых люди зарегистрировались и ввели данные, а затем назначить место для команды.

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

Приведенный выше код выводит следующие результаты

position    miles    team_name
2           134.05   team 1
1           78.00    team 2

Мне бы хотелось, чтобы позиция 1 была назначена команде с наивысшими милями, позиция 2 - второй наивысшей команде ... и т. Д.

Answer 1

В MySQL 8+ вы просто используете 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;

Более ранние версии MySQL поддерживают переменные, но они не очень хорошо работают с GROUP BY и ORDER BY.Решение представляет собой подзапрос (как указано выше):

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

Это работает для меня.

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;