ORACLES的MySQL等价物rank()

Question

Oracle有两个函数-- rank()和dense_rank() --我发现这两个函数对某些应用程序非常有用。我现在正在用mysql做一些事情，我想知道他们是否有类似的东西？

Answer 1

没有直接等价的东西，但是你可以通过一些(不是非常有效的)自连接来伪造它。collection of MySQL query howtos中的一些示例代码

SELECT v1.name, v1.votes, COUNT(v2.votes) AS Rank
FROM votes v1
JOIN votes v2 ON v1.votes < v2.votes OR (v1.votes=v2.votes and v1.name = v2.name)
GROUP BY v1.name, v1.votes
ORDER BY v1.votes DESC, v1.name DESC;
+-------+-------+------+
| name  | votes | Rank |
+-------+-------+------+
| Green |    50 |    1 |
| Black |    40 |    2 |
| White |    20 |    3 |
| Brown |    20 |    3 |
| Jones |    15 |    5 |
| Smith |    10 |    6 |
+-------+-------+------+ 

Answer 2

MySQL中的这个"dense_rank实现“怎么样？

CREATE TABLE `person` (  
 `id` int(11) DEFAULT NULL,  
 `first_name` varchar(20) DEFAULT NULL,  
 `age` int(11) DEFAULT NULL,  
 `gender` char(1) DEFAULT NULL);

INSERT INTO `person` VALUES 
(1,'Bob',25,'M'),
(2,'Jane',20,'F'),
(3,'Jack',30,'M'),
(4,'Bill',32,'M'),
(5,'Nick',22,'M'),
(6,'Kathy',18,'F'),
(7,'Steve',36,'M'),
(8,'Anne',25,'F'),
(9,'Mike',25,'M');

dense_rank()之前的数据如下所示

mysql> select * from person;
+------+------------+------+--------+
| id   | first_name | age  | gender |
+------+------------+------+--------+
|    1 | Bob        |   25 | M      |
|    2 | Jane       |   20 | F      |
|    3 | Jack       |   30 | M      |
|    4 | Bill       |   32 | M      |
|    5 | Nick       |   22 | M      |
|    6 | Kathy      |   18 | F      |
|    7 | Steve      |   36 | M      |
|    8 | Anne       |   25 | F      |
|    9 | Mike       |   25 | M      |
+------+------------+------+--------+
9 rows in set (0.00 sec)

dense_rank()之后的数据如下，包括"partition by“函数

+------------+--------+------+------+
| first_name | gender | age  | rank |
+------------+--------+------+------+
| Anne       | F      |   25 |    1 |
| Jane       | F      |   20 |    2 |
| Kathy      | F      |   18 |    3 |
| Steve      | M      |   36 |    1 |
| Bill       | M      |   32 |    2 |
| Jack       | M      |   30 |    3 |
| Mike       | M      |   25 |    4 |
| Bob        | M      |   25 |    4 |
| Nick       | M      |   22 |    6 |
+------------+--------+------+------+
9 rows in set (0.00 sec)

查询语句为

select first_name,t1.gender,age,FIND_IN_SET(age,t1.age_set) as rank from person t2,
(select gender,group_concat(age order by age desc) as age_set from person group by gender) t1
where t1.gender=t2.gender
order by t1.gender,rank