這幾天開發提交了幾個排名的sql，oracle環境下這類問題就很好解決了，row_number()，rank()或者dense()函數就能搞定，但mysql環境下沒有這類函數，那就自己搞：

測試如下：

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進行排序，grp相同的情況下。我要占位處理：

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相同時我不需要占位，則可以：

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)

更多情況下我需要按照grp分組，然后按照id排序后給出每行的排名，

同樣，當grp相同需要占位時，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)

當grp相同不需要占位時，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)

當然，你可以根據你的需求替換grp和id字段，甚至可以根據自己排名的方式(我這里是正序，你可以倒序)，只是將""就行啦。

總結

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