Question

如果我有三列：

id，username，time

我的数据是：

+-------+------------------+-------------+
|   id  |     username     |    time     |
+-------+------------------+-------------+
|   1   |        A         |    1 min    |
|   2   |        A         |    2 min    |
|   3   |        B         |    3 min    |
|   4   |        B         |    4 min    |
+-------+------------------+-------------+

此查询正在努力获得排名：

   SELECT time, 
    FIND_IN_SET(MIN(time), (SELECT GROUP_CONCAT(time ORDER BY time ASC) 
    FROM table t1)) AS rank
    FROM table t2
    WHERE t2.username = 'B';

只有一个问题：它返回用户B的Rank 3de而不是2nd。

所以我尝试使用 GROUP BY t2.username 以及区分t2.username 但是没有用。

如何获得用户B的排名？它应该是2（不是3 ），因为我们只有2个用户。

Answer 1

E.g：

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(id  INT NOT NULL
,username CHAR(1) NOT NULL
,time TIME NOT NULL
);

INSERT INTO my_table VALUES
(1,'A','00:01:00'),
(2,'A','00:02:00'),
(3,'B','00:03:00'),
(4,'B','00:04:00');

SELECT * FROM my_table;
+----+----------+----------+
| id | username | time     |
+----+----------+----------+
|  1 | A        | 00:01:00 |
|  2 | A        | 00:02:00 |
|  3 | B        | 00:03:00 |
|  4 | B        | 00:04:00 |
+----+----------+----------+

SELECT * 
  FROM 
     ( SELECT username
            , time
            , @i:=@i+1 rank 
         FROM 
            ( SELECT username
                   , MIN(time) time 
                FROM my_table 
               GROUP 
                  BY username
            ) x
            , (SELECT @i:=0) vars 
        ORDER 
           BY time
     ) n 
 WHERE username = 'B';
+----------+----------+------+
| username | time     | rank |
+----------+----------+------+
| B        | 00:03:00 |    2 |
+----------+----------+------+

我认为这也会有用，但它有些笨拙，所以我不确定......

SELECT x.*
     , FIND_IN_SET(time,(SELECT GROUP_CONCAT(DISTINCT time ORDER BY time) FROM (SELECT MIN(time) time FROM my_table GROUP BY username) j )) rank
  FROM my_table x HAVING rank <> 0 AND username = 'B';

独特用户的排名

1 个答案: