Question

让我们假设这个架构：

CREATE TABLE test
(
test_Id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
user_Id INT NOT NULL,
date DATE,
result VARCHAR(255) NOT NULL,
) engine=innodb;

我的目标是为每个不同的 user_Id 选取最后5个结果，从最新到最旧排序。除此之外，根据结果列，我想计算最后结果的比例，以便能够以最佳比率获得3个用户。

因此，让我们以此数据为例：

test_Id | user_Id | date      | result
1       | 1       |2016-09-05 | A
2       | 3       |2016-09-13 | A
3       | 3       |2016-09-30 | A
4       | 4       |2016-09-22 | A
5       | 4       |2016-09-11 | C
6       | 7       |2016-09-18 | D
7       | 4       |2016-09-08 | B
8       | 6       |2016-09-20 | E
9       | 7       |2016-09-16 | A
10      | 7       |2016-09-29 | E
11      | 7       |2016-09-23 | A
12      | 7       |2016-09-16 | B
13      | 4       |2016-09-15 | B
14      | 7       |2016-09-07 | C
15      | 7       |2016-09-09 | A
16      | 3       |2016-09-26 | A
17      | 4       |2016-09-11 | C
18      | 4       |2016-09-30 | E

我能够实现的是这个查询：

SELECT p.user_Id, p.RowNumber, p.date, p.result, 
      SUM(CASE WHEN p.result='A' OR p.result='B' 
      THEN 1 ELSE 0 END) as avg
FROM (
    SELECT  @row_num := IF(@prev_value=user_Id,@row_num+1,1) 
          AS RowNumber, test_Id, user_Id, date, result, 
          @prev_value := user_Id
    FROM test,
    (SELECT @row_num := 1) x,
    (SELECT @prev_value := '') y
    WHERE @prev_value < 5
    ORDER BY user_Id, YEAR(date) DESC, MONTH(date) DESC, 
             DAY(date) DESC
) p
WHERE p.RowNumber <=10
GROUP BY p.user_Id, p.test_Id
ORDER BY p.user_Id, p.RowNumber;

此查询为我提供了这种输出：

 RowNumber |test_Id | user_Id | date      | result | avg
 1         | 1      | 1       |2016-09-05 | A      | 1
 1         | 3      | 3       |2016-09-30 | A      | 1
 2         | 16     | 3       |2016-09-26 | A      | 1
 3         | 2      | 3       |2016-09-13 | A      | 1
 1         | 18     | 4       |2016-09-30 | E      | 0
 2         | 4      | 4       |2016-09-22 | A      | 1
 3         | 13     | 4       |2016-09-15 | B      | 1
 4         | 5      | 4       |2016-09-11 | C      | 0
 5         | 17     | 4       |2016-09-11 | C      | 0
 1         | 8      | 6       |2016-09-20 | E      | 0
 1         | 10     | 7       |2016-09-29 | E      | 0
 2         | 11     | 7       |2016-09-23 | A      | 1
 3         | 6      | 7       |2016-09-18 | D      | 0
 4         | 9      | 7       |2016-09-16 | A      | 1
 5         | 12     | 7       |2016-09-16 | B      | 1

我期待的是，在avg列中，将获得每个用户匹配条件（A或B值）的结果总数，以便能够计算每个user_id的5个结果的比率。（0,0.2,0.4,0.6,0.8,1）。像这样：

 RowNumber |test_Id | user_Id | date      | result | avg
 1         | 1      | 1       |2016-09-05 | A      | 1
 1         | 3      | 3       |2016-09-30 | A      | 3
 2         | 16     | 3       |2016-09-26 | A      | 3
 3         | 2      | 3       |2016-09-13 | A      | 3
 1         | 18     | 4       |2016-09-30 | E      | 2
 2         | 4      | 4       |2016-09-22 | A      | 2
 3         | 13     | 4       |2016-09-15 | B      | 2
 4         | 5      | 4       |2016-09-11 | C      | 2
 5         | 17     | 4       |2016-09-11 | C      | 2
 1         | 8      | 6       |2016-09-20 | E      | 0
 1         | 10     | 7       |2016-09-29 | E      | 3
 2         | 11     | 7       |2016-09-23 | A      | 3
 3         | 6      | 7       |2016-09-18 | D      | 3
 4         | 9      | 7       |2016-09-16 | A      | 3
 5         | 12     | 7       |2016-09-16 | B      | 3

在执行SUM时，我是否受到 GROUP BY p.user_Id，p.test_Id 子句的限制？我只使用user_Id作为GROUP BY子句，并且只将test_Id作为GROUP BY子句尝试了查询，但没有得到预期的结果。

Answer 1

我认为您需要计算平均值然后加入

select  a.rn,a.test_id,a.user_id,a.date,a.result,u.avg from
(
select      t1.*
        ,   if (t1.user_id <> @p, @rn:=1,@rn:=@rn+1) rn
        ,   @p:=t1.user_id p
from        (select @rn:=0, @p:='') rn,test t1
order   by t1.user_id, t1.date desc
) a 
join
(
select  s.user_id
            , sum(case when s.result  = 'A' or s.result = 'B' then 1 else 0 end) as avg
from
(
select      t1.*
        ,   if (t1.user_id <> @p, @rn:=1,@rn:=@rn+1) rn
        ,   @p:=t1.user_id p

from        (select @rn:=0, @p:='') rn,test t1
order   by t1.user_id, t1.date desc
) s
where s.rn <= 5
group   by s.user_id 
) u on u.user_id = a.user_id
where   a.rn <= 5

来自N子查询结果的SUM结果为每个用户的最大值

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