多对多关系示例

Question

我在这里和google中都没有找到任何MYSQL多对多关系的例子。我正在寻找的是看到一个非常简单的例子与php+mysql显示数据库的结果。有人能写一个非常简单的例子吗？

Answer 1

示例场景:一所大学的学生和课程。一个给定的学生可能会上几门课，而且一门课通常会有很多学生。

示例表格，简单的设计：

CREATE TABLE `Student` (
    `StudentID` INT UNSIGNED NOT NULL AUTO_INCREMENT,
    `FirstName` VARCHAR(25),
    `LastName` VARCHAR(25) NOT NULL,
    PRIMARY KEY (`StudentID`)
) ENGINE=INNODB CHARACTER SET utf8 COLLATE utf8_general_ci

CREATE TABLE `Course` (
    `CourseID` SMALLINT UNSIGNED NOT NULL AUTO_INCREMENT,
    `Code` VARCHAR(10) CHARACTER SET ascii COLLATE ascii_general_ci NOT NULL,
    `Name` VARCHAR(100) NOT NULL,
    PRIMARY KEY (`CourseID`)
) ENGINE=INNODB CHARACTER SET utf8 COLLATE utf8_general_ci

CREATE TABLE `CourseMembership` (
    `Student` INT UNSIGNED NOT NULL,
    `Course` SMALLINT UNSIGNED NOT NULL,
    PRIMARY KEY (`Student`, `Course`),
    CONSTRAINT `Constr_CourseMembership_Student_fk`
        FOREIGN KEY `Student_fk` (`Student`) REFERENCES `Student` (`StudentID`)
        ON DELETE CASCADE ON UPDATE CASCADE,
    CONSTRAINT `Constr_CourseMembership_Course_fk`
        FOREIGN KEY `Course_fk` (`Course`) REFERENCES `Course` (`CourseID`)
        ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=INNODB CHARACTER SET ascii COLLATE ascii_general_ci

查找某个课程的所有注册学生：

SELECT
    `Student`.*
FROM
    `Student`
    JOIN `CourseMembership` ON `Student`.`StudentID` = `CourseMembership`.`Student`
WHERE
    `CourseMembership`.`Course` = 1234

查找给定学生选修的所有课程：

SELECT
    `Course`.*
FROM
    `Course`
    JOIN `CourseMembership` ON `Course`.`CourseID` = `CourseMembership`.`Course`
WHERE
    `CourseMembership`.`Student` = 5678

Answer 2

这里有一个涉及到的SQL的快速而肮脏的示例。我不认为有任何必要用php来混淆这个概念。只需像检索其他集合一样检索集合即可。

在本例中，有许多名称和颜色。人们可以有多个喜欢的颜色，许多人可以有相同的喜欢的颜色。因此，多对多。

***** Tables **********

person
--------
id - int 
name - varchar

favColor
-------------
id - int 
color - varchar

person_color
------------
person_id - int (matches an id from person)
color_id - int (matches an id from favColor)



****** Sample Query ******

SELECT name, color 
FROM person 
    LEFT JOIN person_color ON (person.id=person_id)
    LEFT JOIN favColor ON (favColor.id=color_id)


****** Results From Sample Query *******

Name - Color
---------------
John - Blue
John - Red
Mary - Yellow
Timmy - Yellow
Suzie - Green
Suzie - Blue
etc...

这有帮助吗？

Answer 3

mysql> SELECT * FROm products;
+----+-----------+------------+
| id | name      | company_id |
+----+-----------+------------+
|  1 | grechka   |          1 |
|  2 | rus       |          1 |
|  3 | makaronu  |          2 |
|  4 | yachna    |          3 |
|  5 | svuniacha |          3 |
|  6 | manka     |          4 |
+----+-----------+------------+
6 rows in set (0.00 sec)

mysql> SELECT * FROm company;
+----+----------+
| id | name     |
+----+----------+
|  1 | LVIV     |
|  2 | KIEV     |
|  3 | KHarkiv  |
|  4 | MADRID   |
|  5 | MILAN   |
|  6 | KOR |
+----+----------+
6 rows in set (0.00 sec)

mysql> SELECT * FROm many_many;
+------------+---------+
| product_id | city_id |
+------------+---------+
|          1 |       1 |
|          1 |       3 |
|          2 |       3 |
|          1 |       2 |
|          1 |       4 |
|          2 |       4 |
|          2 |       1 |
|          3 |       1 |
+------------+---------+
8 rows in set (0.00 sec)

mysql> SELECT products.name,company.name FROM products JOIN many_many ON many_
ny.product_id =products.id JOIN company ON company.id= many_many.city_id;
+----------+---------+
| name     | name    |
+----------+---------+
| grechka  | LVIV    |
| grechka  | KHarkiv |
| grechka  | KIEV    |
| grechka  | MADRID  |
| rus      | KHarkiv |
| rus      | MADRID  |
| rus      | LVIV    |
| makaronu | LVIV    |
+----------+---------+
8 rows in set (0.00 sec)