Eksempler på mange-til-mange forhold

Eksempelscenarie:studerende og kurser på et universitet. En given elev kan være på flere kurser, og naturligvis vil et kursus normalt have mange studerende.

Eksempel tabeller, enkelt design:

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

Find alle elever, der er tilmeldt et kursus:

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

Find alle kurser taget af en given studerende:

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