增强 SQL 架构

Question

我需要设计一个没有重复组的 table，我想我已经做到了，但是 table 看起来与它们的链接方式等是否正确，或者有什么我可以做的在...上加强？我运行所有的查询，我有几个重复的球员和经理组，因为我把他们都放在一个小队 table 中，所以我把它们分成 team_player 和 team_manager 和这样一个或多个经理也可以管理一个团队，反之亦然管理体育场等

drop table film_director;
drop table film_actor;
drop table film;
drop table studio;
drop table actor;
drop table director;

CREATE TABLE studio(
  studio_ID NUMBER NOT NULL,
  studio_Name VARCHAR2(30),
  PRIMARY KEY(studio_ID));

CREATE TABLE film(
  film_ID NUMBER NOT NULL,
  studio_ID NUMBER NOT NULL,
  genre VARCHAR2(30),
  genre_ID NUMBER(1),
  film_Len NUMBER(3),
  film_Title VARCHAR2(30) NOT NULL,
  year_Released NUMBER NOT NULL,
  PRIMARY KEY(film_ID),
  FOREIGN KEY (studio_ID) REFERENCES studio);

CREATE TABLE director(
  director_ID NUMBER NOT NULL,
  director_fname VARCHAR2(30),
  director_lname VARCHAR2(30),
  PRIMARY KEY(director_ID));

CREATE TABLE actor(
  actor_ID NUMBER NOT NULL,
  actor_fname VARCHAR2(15),
  actor_lname VARCHAR2(15),
  PRIMARY KEY(actor_ID));

CREATE TABLE film_actor(
  film_ID NUMBER NOT NULL,
  actor_ID NUMBER NOT NULL,
  PRIMARY KEY(film_ID, actor_ID),
  FOREIGN KEY(film_ID) REFERENCES film(film_ID),
  FOREIGN KEY(actor_ID) REFERENCES actor(actor_ID));

CREATE TABLE film_director(
  film_ID NUMBER NOT NULL,
  director_ID NUMBER NOT NULL,
  PRIMARY KEY(film_ID, director_ID),
  FOREIGN KEY(film_ID) REFERENCES film(film_ID),
  FOREIGN KEY(director_ID) REFERENCES director(director_ID));

INSERT INTO studio (studio_ID, studio_Name) VALUES (1, 'Paramount');
INSERT INTO studio (studio_ID, studio_Name) VALUES (2, 'Warner Bros');
INSERT INTO studio (studio_ID, studio_Name) VALUES (3, 'Film4');
INSERT INTO studio (studio_ID, studio_Name) VALUES (4, 'Working Title Films');

INSERT INTO film (film_ID, studio_ID, genre, genre_ID, film_Len, film_Title, year_Released) VALUES (1, 1, 'Comedy', 1, 180, 'The Wolf Of Wall Street', 2013);
INSERT INTO film (film_ID, studio_ID, genre, genre_ID, film_Len, film_Title, year_Released) VALUES (2, 2, 'Romance', 2, 143, 'The Great Gatsby', 2013);
INSERT INTO film (film_ID, studio_ID, genre, genre_ID, film_Len, film_Title, year_Released) VALUES (3, 3, 'Science Fiction', 3, 103, 'Never Let Me Go', 2008);
INSERT INTO film (film_ID, studio_ID, genre, genre_ID, film_Len, film_Title, year_Released) VALUES (4, 4, 'Romance', 4, 127, 'Pride and Prejudice', 2005);

INSERT INTO director (director_ID, director_fname, director_lname) VALUES (1, 'Martin', 'Scorcese');
INSERT INTO director (director_ID, director_fname, director_lname) VALUES (2, 'Baz', 'Luhrmann');
INSERT INTO director (director_ID, director_fname, director_lname) VALUES (3, 'Mark', 'Romanek');
INSERT INTO director (director_ID, director_fname, director_lname) VALUES (4, 'Joe', 'Wright');

INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (1, 'Matthew', 'McConnaughy');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (2, 'Leonardo', 'DiCaprio');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (3, 'Margot', 'Robbie');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (4, 'Joanna', 'Lumley');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (5, 'Carey', 'Mulligan');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (6, 'Tobey', 'Maguire');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (7, 'Joel', 'Edgerton');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (8, 'Keira', 'Knightly');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (9, 'Andrew', 'Garfield');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (10, 'Sally', 'Hawkins');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (11, 'Judi', 'Dench');
INSERT INTO actor (actor_ID, actor_fname, actor_lname) VALUES (12, 'Matthew', 'Macfadyen');

INSERT INTO film_actor (film_ID, actor_ID) VALUES (1, 1);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (1, 2);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (1, 3);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (1, 4);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (2, 2);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (2, 5);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (2, 6);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (2, 7);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (3, 5);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (3, 8);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (3, 9);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (3, 10);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (4, 5);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (4, 8);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (4, 11);
INSERT INTO film_actor (film_ID, actor_ID) VALUES (4, 12);

INSERT INTO film_director (film_ID, director_ID) VALUES (1,1);
INSERT INTO film_director (film_ID, director_ID) VALUES (2,2);
INSERT INTO film_director (film_ID, director_ID) VALUES (3,3);
INSERT INTO film_director (film_ID, director_ID) VALUES (4,4);

Answer 1

我只会为用户提供 1 个 table (director/actor)，因为如果导演也是像昆汀·塔伦蒂诺 (Quentin Tarantino) 这样的演员，他会出现在他的大部分电影中。您将不得不重复数据。

https://en.wikipedia.org/wiki/Database_normalization

尝试将您的模型提升到 BCNF 或 4NF，您会没事的

Answer 2

我同意 sdmon - 因为导演和演员具有相同的属性，所以将他们全部放在一个 table 中并在 "facts" table 中为他们分配角色是有意义的]s.

一部电影可以有多个导演吗？如果没有，那么你不需要 film_director table，你只需要电影 table 中的 "director" 列。否则你确实需要 film_director table.

一部电影可以绑定到多个工作室吗？在您的模型中，电影 table 中有一个 studio_id 列，还有一个 film_studio table。你不需要两者。如果一部电影总是由一个工作室制作，则您需要该列（而不是 table）。如果您需要为一部电影放映多个工作室，那么您需要额外的 table，而不是列。

其他一切看起来都很好。祝你好运！

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