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()函式就能搞定，但mysql環境下沒有這類函式，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。我要佔位處理：
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相同時我不需要佔位，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，然後按照id排序後給出每行的排名，
同樣，當grp相同需要佔位時，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相同不需要佔位時，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欄位，甚至可以根據自己排名的方式（我這裡是正序，你可以倒序），只是將""就行啦。

出處:http://mxohy.blog.sohu.com/172181390.html?qq-pf-to=pcqq.c2c

mysql上排名sql的寫法，類似oracle的rank和dense

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