Question

我需要编写一个SQL查询来识别运行时间最长的电影的标题，我只是想知道如何做到这一点？我已经尝试了这个，但我不确定我需要做些什么才能解决这个问题。

select f.film_title
from film f
order by f.film_len desc
limit 1;

我认为最简单的方法是按照长度对电影进行排序，然后按升序排序。然后只取第一个结果，这将是最长的电影。但是，这并没有考虑到长度相同的电影。

这是我创建的表格，我必须从中找到结果。

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

- 您必须假设将有相同运行时的电影。

select f.film_title
from film f
where film_Len = 
(select max(film_Len) from film)

Answer 2

select * from
(select f.film_title
from film f
order by f.film_len desc) as f
where rownum <= 1;

更新适用于订购后的位置。

Answer 3

您还可以使用排名功能来确定：

SELECT *
FROM
  (SELECT f.film_title,
    rank() over(partition BY f.film_title order by f.film_len DESC) rnk
    from film f
  )
WHERE rnk = 1

如果有2部长度相同的电影，则会显示。

Answer 4

您可以使用 pagination query ：

Oracle 11g 以及之前：

SELECT *
FROM
  ( SELECT f.film_title FROM film f ORDER BY f.film_len DESC
  )
WHERE ROWNUM = 1;

Oracle 12c ：前N行限制功能

SELECT f.film_title 
FROM film f 
ORDER BY f.film_len DESC
FETCH FIRST 1 ROW ONLY;

SQL ORDER和LIMIT为1的结果

4 个答案: