数据库设计问题 – SQL

要求：a 表：`id`, `name` ; 作为词表，存放不同的词；b 表：`id`, `attr` ; 作为属性表，存放各种属性；其中，一个词可以有不同的多个属性；而每个词的属性的个数也不一定相同；c 表：`id`, `aid`, `bid` ; 作为关系表，存放每个词的对应关系；写出 SQL 语句，来得到每个词拥有属性总数的逆向（DESC）排序：

各种表的信息如下：

mysql> DESC `a`; DESC `b`; DESC `c`;
+-------+---------------------+------+-----+---------+----------------+
| Field | Type                | Null | Key | Default | Extra          |
+-------+---------------------+------+-----+---------+----------------+
| id    | bigint(20) unsigned | NO   | PRI | NULL    | auto_increment |
| name  | varchar(255)        | NO   |     | NULL    |                |
+-------+---------------------+------+-----+---------+----------------+
2 rows in set (0.00 sec)

+-------+---------------------+------+-----+---------+----------------+
| Field | Type                | Null | Key | Default | Extra          |
+-------+---------------------+------+-----+---------+----------------+
| id    | bigint(20) unsigned | NO   | PRI | NULL    | auto_increment |
| attr  | varchar(255)        | NO   |     | NULL    |                |
+-------+---------------------+------+-----+---------+----------------+
2 rows in set (0.01 sec)

+-------+---------------------+------+-----+---------+----------------+
| Field | Type                | Null | Key | Default | Extra          |
+-------+---------------------+------+-----+---------+----------------+
| id    | bigint(20) unsigned | NO   | PRI | NULL    | auto_increment |
| aid   | int(8)              | NO   |     | NULL    |                |
| bid   | int(8)              | NO   |     | NULL    |                |
+-------+---------------------+------+-----+---------+----------------+
3 rows in set (0.00 sec)

我们预先放入测试的数据，如下：

mysql> SELECT * FROM `a`; SELECT * FROM `b`; SELECT * FROM `c`;
+----+------+
| id | name |
+----+------+
|  1 | a    |
|  2 | b    |
|  3 | c    |
|  4 | d    |
|  5 | e    |
+----+------+
5 rows in set (0.00 sec)

+----+------+
| id | attr |
+----+------+
|  1 | 111  |
|  2 | 112  |
|  3 | 113  |
|  4 | 123  |
|  5 | 221  |
|  6 | 231  |
|  7 | 252  |
|  8 | 278  |
|  9 | 292  |
| 10 | 256  |
| 11 | 578  |
| 12 | 653  |
| 13 | 521  |
| 14 | 502  |
+----+------+
14 rows in set (0.00 sec)

+----+-----+-----+
| id | aid | bid |
+----+-----+-----+
|  1 |   1 |   1 |
|  2 |   1 |   2 |
|  3 |   1 |   4 |
|  4 |   1 |   7 |
|  5 |   2 |   8 |
|  6 |   2 |  11 |
|  7 |   3 |   3 |
|  8 |   3 |   5 |
|  9 |   3 |   6 |
| 10 |   4 |   9 |
| 11 |   4 |  10 |
| 12 |   5 |  12 |
| 13 |   5 |  13 |
| 14 |   5 |  14 |
+----+-----+-----+
14 rows in set (0.00 sec)

首先执行下列语句：

mysql> SELECT COUNT(`bid`) AS `attrcounts` FROM `c` GROUP BY `aid` ORDER BY `attrcounts` DESC;
+------------+
| attrcounts |
+------------+
|          4 |
|          3 |
|          3 |
|          2 |
|          2 |
+------------+
5 rows in set (0.00 sec)

进而，我们再连表：

mysql> SELECT a.name, COUNT(c.bid) AS `attrcounts` FROM `c` LEFT JOIN `a` ON a.id = c.aid GROUP BY c.aid ORDER BY `attrcounts` DESC;
+------+------------+
| name | attrcounts |
+------+------------+
| a    |          4 |
| c    |          3 |
| e    |          3 |
| b    |          2 |
| d    |          2 |
+------+------------+
5 rows in set (0.00 sec)

于是，我们得到了结果；

-------

补充一些基础知识：

如何修改已有表的列：http://www.w3school.com.cn/sql/sql_alter.asp

GROUP BY 相关知识：http://www.w3school.com.cn/sql/sql_groupby.asp