mysql查询排名

测试如下：

mysql> select * from animals_inno;

+--------+----+------------+---------------------+----------+

| grp    | id | name       | created             | modified |

+--------+----+------------+---------------------+----------+

| mammal |  1 | dog        | 0000-00-00 00:00:00 | NULL     |

| mammal |  2 | cat        | 0000-00-00 00:00:00 | NULL     |

| bird   |  3 | penguin    | 0000-00-00 00:00:00 | NULL     |

| fish   |  4 | lax        | 0000-00-00 00:00:00 | NULL     |

| mammal |  5 | whale      | 0000-00-00 00:00:00 | NULL     |

| bird   |  6 | ?????????? | 2011-04-13 14:52:48 | NULL     |

| bird   |  7 | ostrich    | 0000-00-00 00:00:00 | NULL     |

| fish   |  8 |            | 0000-00-00 00:00:00 | NULL     |

| fish   |  9 | NULL       | 0000-00-00 00:00:00 | NULL     |

+--------+----+------------+---------------------+----------+

9 rows in set (0.00 sec)

我想要按照grp进行排序，grp相同的情况下。我要占位处理：

SELECT grp,

       name,

       id,

       (SELECT COUNT(*) FROM animals_inno where grp < a.grp) + 1 place

  FROM animals_inno a

 ORDER BY place;

+--------+------------+----+-------+

| grp    | name       | id | place |

+--------+------------+----+-------+

| bird   | penguin    |  3 |     1 |

| bird   | ?????????? |  6 |     1 |

| bird   | ostrich    |  7 |     1 |

| fish   | lax        |  4 |     4 |

| fish   |            |  8 |     4 |

| fish   | NULL       |  9 |     4 |

| mammal | dog        |  1 |     7 |

| mammal | cat        |  2 |     7 |

| mammal | whale      |  5 |     7 |

+--------+------------+----+-------+

9 rows in set (0.00 sec)

如果grp相同时我不需要占位，则可以：

select grp,

       name,

       id,

       (select count(distinct grp) from animals_inno where grp < a.grp) + 1 place

  from animals_inno a

 order by place;

+--------+------------+----+-------+

| grp    | name       | id | place |

+--------+------------+----+-------+

| bird   | penguin    |  3 |     1 |

| bird   | ?????????? |  6 |     1 |

| bird   | ostrich    |  7 |     1 |

| fish   | lax        |  4 |     2 |

| fish   |            |  8 |     2 |

| fish   | NULL       |  9 |     2 |

| mammal | dog        |  1 |     3 |

| mammal | cat        |  2 |     3 |

| mammal | whale      |  5 |     3 |

+--------+------------+----+-------+

9 rows in set (0.00 sec)

更多情况下我需要按照grp分组，然后按照id排序后给出每行的排名，

同样，当grp相同需要占位时，可以：

SELECT grp,

       name,

       id,

       (SELECT COUNT(*) FROM animals_inno where grp =a.grp and id< a.id) + 1 place

  FROM animals_inno a

 ORDER BY grp,place;

+--------+------------+----+-------+

| grp    | name       | id | place |

+--------+------------+----+-------+

| fish   | lax        |  4 |     1 |

| fish   |            |  8 |     2 |

| fish   | NULL       |  9 |     3 |

| mammal | dog        |  1 |     1 |

| mammal | cat        |  2 |     2 |

| mammal | whale      |  5 |     3 |

| bird   | penguin    |  3 |     1 |

| bird   | ?????????? |  6 |     2 |

| bird   | ostrich    |  7 |     3 |

+--------+------------+----+-------+

9 rows in set (0.00 sec)

当grp相同不需要占位时，可以：

SELECT grp,

       name,

       id,

       (SELECT COUNT(distinct id) FROM animals_inno where grp =a.grp and id< a.id) + 1 place

  FROM animals_inno a

 ORDER BY grp,place;

+--------+------------+----+-------+

| grp    | name       | id | place |

+--------+------------+----+-------+

| fish   | lax        |  4 |     1 |

| fish   |            |  8 |     2 |

| fish   | NULL       |  9 |     3 |

| mammal | dog        |  1 |     1 |

| mammal | cat        |  2 |     2 |

| mammal | whale      |  5 |     3 |

| bird   | penguin    |  3 |     1 |

| bird   | ?????????? |  6 |     2 |

| bird   | ostrich    |  7 |     3 |

+--------+------------+----+-------+

9 rows in set (0.00 sec)