MAX（）和GROUP BY的相关子查询

Question

我有使用MAX（）和GROUP BY的问题。 我旁边表：

personal_prizes

 
___________ ___________ _________ __________ 
| id  | userId | specId| group | 
|___________|___________|_________|__________| 
| 1  | 1  | 1  |  1 | 
|___________|___________|_________|__________| 
| 2  | 1  | 2  |  1 | 
|___________|___________|_________|__________| 
| 3  | 2  | 3  |  1 | 
|___________|___________|_________|__________| 
|  4  | 2  | 4  |  2 | 
|___________|___________|_________|__________| 
|  5  | 1  | 5  |  2 | 
|___________|___________|_________|__________| 
|  6  | 1  | 6  |  2 | 
|___________|___________|_________|__________| 
| 7  | 2  | 7  |  3 | 
|___________|___________|_________|__________| 


prizes 
___________ ___________ _________ 
| id  | title | group |  
|___________|___________|_________| 
| 1  | First | 1  | 
|___________|___________|_________| 
| 2  | Second | 1  | 
|___________|___________|_________| 
| 3  | Newby | 1  | 
|___________|___________|_________| 
|  4  | General| 2  | 
|___________|___________|_________| 
|  5  | Leter | 2  | 
|___________|___________|_________| 
|  6  | Ter | 2  | 
|___________|___________|_________| 
|  7  | Mentor | 3  | 
|___________|___________|_________| 

所以，我需要选择用户最高头衔。 例如id = 1的用户必须有奖品'Second'，'Ter'。 我不知道如何实现它在一个查询（（（ 所以，首先，我尽量选择用户最高规范ID 接下来我尝试：

SELECT pp.specID 
FROM personal_prizes pp 
WHERE pp.specID IN (SELECT MAX(pp1.id) 
        FROM personal_prizes pp1 
        WHERE pp1.userId = 1 
        GROUP BY pp1.group) 

而且它不工作 。所以请帮我解决这个问题 如果你帮助为用户挑选奖品，它将会很棒！

Answer 1

我在这里看到的问题是，prizes.id确实不是一个确定哪个是可靠的方法“最高”的奖金，但忽略这一点，我建议使用ROW_NUMBER（）OVER（）来定位每个用户的“最高”奖励，如下所示：

请参阅本SQL Fiddle

CREATE TABLE personal_prizes 
    ([id] int, [userId] int, [specId] int, [group] int) 
; 

INSERT INTO personal_prizes 
    ([id], [userId], [specId], [group]) 
VALUES 
    (1, 1, 1, 1), 
    (2, 1, 2, 1), 
    (3, 2, 3, 1), 
    (4, 2, 4, 2), 
    (5, 1, 5, 2), 
    (6, 1, 6, 2), 
    (7, 2, 7, 3) 
; 


CREATE TABLE prizes 
    ([id] int, [title] varchar(7), [group] int) 
; 

INSERT INTO prizes 
    ([id], [title], [group]) 
VALUES 
    (1, 'First', 1), 
    (2, 'Second', 1), 
    (3, 'Newby', 1), 
    (4, 'General', 2), 
    (5, 'Leter', 2), 
    (6, 'Ter', 2), 
    (7, 'Mentor', 3) 
; 

查询1：

select 
* 
from (
    select 
     pp.*, p.title 
    , row_number() over(partition by pp.userId order by p.id ASC) as prize_order 
    from personal_prizes pp 
    inner join prizes p on pp.specid = p.id 
    ) d 
where prize_order = 1 

Results：

| id | userId | specId | group | title | prize_order | 
|----|--------|--------|-------|-------|-------------| 
| 1 |  1 |  1 |  1 | First |   1 | 
| 3 |  2 |  3 |  1 | Newby |   1 | 

通过改变ORDER BY内的结果可以 “反向” o版本条款：

select 
* 
from (
    select 
     pp.*, p.title 
    , row_number() over(partition by pp.userId order by p.id DESC) as prize_order 
    from personal_prizes pp 
    inner join prizes p on pp.specid = p.id 
    ) d 
where prize_order = 1 

| id | userId | specId | group | title | prize_order | 
|----|--------|--------|-------|--------|-------------| 
| 6 |  1 |  6 |  2 | Ter |   1 | 
| 7 |  2 |  7 |  3 | Mentor |   1 | 

你可以在这个逻辑扩展到找到“每组最高奖”太

select 
* 
from (
    select 
     pp.*, p.title 
    , row_number() over(partition by pp.userId, p.[group] order by p.id ASC) as prize_order 
    from personal_prizes pp 
    inner join prizes p on pp.specid = p.id 
    ) d 
where prize_order = 1 


| id | userId | specId | group | title | prize_order | 
|----|--------|--------|-------|---------|-------------| 
| 1 |  1 |  1 |  1 | First |   1 | 
| 5 |  1 |  5 |  2 | Leter |   1 | 
| 3 |  2 |  3 |  1 | Newby |   1 | 
| 4 |  2 |  4 |  2 | General |   1 | 
| 7 |  2 |  7 |  3 | Mentor |   1 |