Question

我有以下活动表。

| user_id | create_date | start_time | field_1 | field_2 | field_3 |    
|       1 | 2017-03-01  |          0 |       1 |       1 |       1 |  
|       1 | 2017-03-01  |         30 |       2 |       2 |       2 |  
|       1 | 2017-03-01  |         60 |       3 |       3 |       3 |  
|       1 | 2017-03-02  |          0 |       4 |       4 |       4 |  
|       1 | 2017-03-02  |         30 |       5 |       5 |       5 |  
|       1 | 2017-03-02  |         60 |       6 |       6 |       6 |  
|       1 | 2017-03-03  |          0 |       7 |       7 |       7 |  
|       1 | 2017-03-03  |         30 |       8 |       8 |       8 |  
|       1 | 2017-03-03  |         60 |       9 |       9 |       9 | 
|       2 | 2017-03-04  |          0 |       1 |       1 |       1 |  
|       2 | 2017-03-04  |         30 |       2 |       2 |       2 |  
|       2 | 2017-03-04  |         60 |       3 |       3 |       3 |  
|       2 | 2017-03-05  |          0 |       4 |       4 |       4 |  
|       2 | 2017-03-05  |         30 |       5 |       5 |       5 |  
|       2 | 2017-03-05  |         60 |       6 |       6 |       6 |  
|       2 | 2017-03-06  |          0 |       7 |       7 |       7 |  
|       2 | 2017-03-06  |         30 |       8 |       8 |       8 |  
|       2 | 2017-03-06  |         60 |       9 |       9 |       9 |

从活动中选择*;

| user_id | create_date | best_active_time |  
|       1 | 2017-03-03  | 72               |  
|       2 | 2017-03-06  | 72               |

我想要的内容适用于每个user_id，即field_1，field_2，field_3之和最大的日期。在上面的示例中，查询结果应为

SELECT bb.user_id, bb.create_date, cc.best_active_time FROM   
(  
SELECT user_id, create_date, MAX(active_time) best_active_time FROM  
(SELECT user_id, create_date, (SUM(field_1) + SUM(field_2) + SUM(field_3)) as   active_time  FROM activity GROUP BY user_id, create_date) ACT GROUP BY user_id, create_date  
) bb  
INNER JOIN  
(  
SELECT user_id,  MAX(active_time) best_active_time FROM  
(SELECT user_id, (SUM(field_1) + SUM(field_2) + SUM(field_3)) as active_time    FROM activity GROUP BY user_id, create_date) ACT GROUP BY user_id  
) cc ON bb.user_id = cc.user_id AND bb.best_active_time = cc.best_active_time

该问题与获取非聚合列（create_date）和使用GROUP BY子句有关。目前，我有我的解决方案，我只是想知道是否存在更好的解决方案（我觉得它们存在）。

function Troop(armor, weapon, shield) {
            var armor = armor;
            var weapon = weapon;
            var shield = shield;

            this.getTroopDetails = function() {
                console.log("Armour -> " + armor + ", Weapon -> " + weapon + ", Shield -> " + shield);
            }
        }

Answer 1

您可以使用变量在mysql中模拟row_number()。

<强> SQL DEMO

SELECT *
FROM (
        SELECT a.*,
               @rn := if(@user = user_id, 
                         @rn + 1 ,
                         if(@user := user_id,1,1)
                        ) as rn
        FROM (        
                SELECT user_id, create_date, SUM(field_1 + field_2 + field_3) as total
                FROM activity a
                GROUP BY user_id, create_date
             ) a
        CROSS JOIN (SELECT @user := 0, @rn := 0) as t
        ORDER BY user_id, total DESC
    ) t
WHERE rn =1

<强>输出：

非聚合列，SUM，MAX和GROUP BY

1 个答案: