Question

我有一个具有这种结构的表：

score_id | user_id | category | subcategory | score
1        | 1       | game     | 0           | 100
2        | 1       | game     | 0           | 200
3        | 1       | quiz     | 0           | 2000
4        | 1       | quiz     | 1           | 1000
5        | 1       | game     | 0           | 10
6        | 1       | game     | 1           | 10
7        | 1       | game     | 2           | 100
8        | 1       | game     | 1           | 500
9        | 2       | game     | 0           | 600

我需要不同的查询mysql：

1）所有团队的排名按user_id分类所有最佳结果的总和，用户类别和子类别的所有记录

预期结果

user_id | total_score | ranking
1       | 3310        | 1
2       | 600         | 2

，其中

3310 = (200 + 10 + 100+1000+2000) 
 200 is the best result of game 0, 
 10 is the best result of game 1, 
 100 is the best result of game 2, 
 2000 is the best result of quiz 0
 1000 is the best result of quiz 1

600 = (600)
 600 is the best result of game 0,

@Strawberry的解决方案（谢谢）

SELECT a.*
     , @i:=@i+1 rank
  FROM 
     ( SELECT user_id
            , SUM(subtotal) total 
         FROM 
           ( SELECT user_id
                   , category
                   , subcategory
                   , MAX(score) subtotal 
                FROM my_table 
               GROUP 
                  BY user_id
                   , category
                   , subcategory
            ) x 
        GROUP 
           BY user_id
     ) a
  JOIN 
     ( SELECT @i:=0) vars
 ORDER
    BY total DESC;

2）所有团队的排名按user_id排序所有最佳结果的总和，用于所有类别和子类别的记录仅用于游戏或测验

预期结果游戏

user_id | total_score | ranking
2       | 600         | 1
1       | 310         | 2

其中

600 = (600)
 600 is the best result of game 0, 

310 = (200 + 10 + 100) 
 200 is the best result of game 0, 
 10 is the best result of game 1, 
 100 is the best result of game 2,

预期结果测验

user_id | total_score | ranking
1       | 3000       | 1
2       | 0          | 2

其中

3000 = (200 + 10 + 100) 
 2000 is the best result of quiz 0
 1000 is the best result of quiz 1

 0= (0)
(user_id= 2 don't play quiz)

基于@Strawberry的解决方案（谢谢）

SELECT a.*
     , @i:=@i+1 rank
  FROM 
     ( SELECT user_id
            , SUM(subtotal) total 
         FROM 
           ( SELECT user_id
                   , category
                   , subcategory
                   , MAX(score) subtotal 
                FROM my_table
                WHERE category = 'game' // or 'quiz
               GROUP 
                  BY user_id
                   , category
                   , subcategory
            ) x 
        GROUP 
           BY user_id
     ) a
  JOIN 
     ( SELECT @i:=0) vars
 ORDER
    BY total DESC;

3）所有团队的排名按user_id排序所有最佳结果的总和，用于所有类别和子类别的用户记录仅适用于具有特定子类别的游戏或测验

预期结果游戏0

user_id | total_score | ranking
2       | 600         | 1
1       | 310         | 2

其中

600 = (600)
 600 is the best result of game 0, 

200 = (200 ) 
 200 is the best result of game 0,

基于@Strawberry的解决方案（谢谢）

SELECT a.*
     , @i:=@i+1 rank
  FROM 
     ( SELECT user_id
            , SUM(subtotal) total 
         FROM 
           ( SELECT user_id
                   , category
                   , subcategory
                   , MAX(score) subtotal 
                FROM my_table
                WHERE category = 'game'
                AND subcategory = '0'
               GROUP 
                  BY user_id
                   , category
                   , subcategory
            ) x 
        GROUP 
           BY user_id
     ) a
  JOIN 
     ( SELECT @i:=0) vars
 ORDER
    BY total DESC;

4）获得查询1的单个用户的总分（ex user_id = 1）

5）获得查询1的单个用户（ex user_id = 1）的排名

6）获得查询2的单个用户的总分（ex user_id = 1）

7）获得查询2的单个用户（ex user_id = 1）的排名

8）获得查询3的单个用户的总分（ex user_id = 1）

9）获得查询3的单个用户（ex user_id = 1）的排名

谢谢你！

Answer 1

这是第一个。使用此功能，请向我们展示您为其余部分所做的最佳努力......

In [184]: s[4:] / s[:-4].values - 1
Out[184]: 
2000-01-01 04:00:00         inf
2000-01-01 05:00:00    4.000000
2000-01-01 06:00:00    2.000000
Freq: H, dtype: float64

请注意，此解决方案不考虑关系。

mysql排名总和和最佳分数

1 个答案: