sqlzoo more join 题目及答案

博客围绕电影数据展开一系列SQL查询,包括查询特定年份电影、演员ID、电影的演员列表等,还涉及按条件筛选和排序,如按年份、演员数量等,使用的数据库相关技术为MySQL和SQL。
  1. List the films where the yr is 1962 [Show id, title]
SELECT id, title
 FROM movie
 WHERE yr=1962
  1. Give year of ‘Citizen Kane’.
select yr
from movie
where title = 'Citizen Kane'
  1. List all of the Star Trek movies, include the id, title and yr (all of these movies include the words Star Trek in the title). Order results by year.
select id, title, yr
from movie
where title like '%Star Trek%'
order by yr
  1. What id number does the actor ‘Glenn Close’ have?
select id
from actor
where name = 'Glenn Close'
  1. What is the id of the film ‘Casablanca’
select id
from movie
where title = 'Casablanca'
  1. Obtain the cast list for ‘Casablanca’.
select a.name
from actor a
join casting c
on a.id = c.actorid
where movieid = 11768
  1. Obtain the cast list for the film ‘Alien’
select name
from actor a
join casting c on a.id = c.actorid
join movie m on m.id = c.movieid
where title = 'Alien'
  1. List the films in which ‘Harrison Ford’ has appeared
select title
from movie m
join casting c on m.id = c.movieid
join actor a on a.id = c.actorid
where name = 'Harrison Ford'
  1. List the films where ‘Harrison Ford’ has appeared - but not in the starring role. [Note: the ord field of casting gives the position of the actor. If ord=1 then this actor is in the starring role]
select title
from movie m
join casting c on m.id = c.movieid
join actor a on a.id = c.actorid
where name = 'Harrison Ford' and c.ord != 1
  1. List the films together with the leading star for all 1962 films.
select title, name
from movie m
join casting c on m.id = c.movieid
join actor a on a.id = c.actorid
where yr = 1962 and c.ord = 1
  1. Which were the busiest years for ‘Rock Hudson’, show the year and the number of movies he made each year for any year in which he made more than 2 movies.
SELECT yr,COUNT(title) FROM
  movie JOIN casting ON movie.id=movieid
        JOIN actor   ON actorid=actor.id
WHERE name='Rock Hudson'
GROUP BY yr
HAVING COUNT(title) > 2
  1. List the film title and the leading actor for all of the films ‘Julie Andrews’ played in.
select title, name
from movie m
join casting c on m.id = c.movieid
join actor a on a.id = c.actorid
where movieid in (
	SELECT movieid 
	FROM casting
	WHERE actorid IN (
  		SELECT id 
  		FROM actor
 		WHERE name='Julie Andrews')
)
and ord = 1
  1. Obtain a list, in alphabetical order, of actors who’ve had at least 15 starring roles.
select name 
from actor
where id in (
	select actorid
	from casting
	where ord = 1
	group by actorid
	having count(*) >= 15
)
order by name
  1. List the films released in the year 1978 ordered by the number of actors in the cast, then by title.
select title, count(actorid)
from movie m 
join casting c on m.id = c.movieid
join actor a on a.id = c.actorid
where yr = 1978
group by title
order by count(actorid) desc, title
  1. List all the people who have worked with ‘Art Garfunkel’.
select name
from actor a
join casting c on a.id = c.actorid
where movieid in (
	select movieid
	from actor a
	join casting c on a.id = c.actorid
	where name = 'Art Garfunkel'
)
and name != 'Art Garfunkel'
### SQLZOO JOIN 练习解答 在处理数据库查询时,`JOIN` 是一种用于组合来自两个或多个表的数据的重要工具。对于 `LEFT JOIN` 和 `RIGHT JOIN` 的区别,在某些情况下,选择合适的连接方式可以极大地影响查询的结果集。 #### 左连接 (LEFT JOIN) 左连接返回左表中的所有记录以及右表中存在的匹配记录。如果右表中不存在匹配,则结果集中相应的列将包含 NULL 值[^1]。 ```sql SELECT A.column_name, B.column_name FROM TableA AS A LEFT JOIN TableB AS B ON A.common_field = B.common_field; ``` #### 右连接 (RIGHT JOIN) 右连接则相反,它会返回右表中的所有记录以及左表中存在的匹配记录。同样地,当左表缺少对应数据时,这些字段会被填充为 NULL。 ```sql SELECT A.column_name, B.column_name FROM TableA AS A RIGHT JOIN TableB AS B ON A.common_field = B.common_field; ``` 为了更好地理解如何应用这两种类型的联接操作,下面提供几个基于 SQLZOO 平台上的练习实例: ##### 示例 1: 查询世界人口超过五千五百万的国家及其所属大洲的人口总和 此问题可以通过执行内连接来解决,因为这里只需要考虑那些确实存在于两张表格内的条目;但是也可以通过其他形式的外键关联实现相同的功能。 ```sql SELECT continent, SUM(population) FROM world WHERE population > 55000000 GROUP BY continent; -- 或者使用 INNER JOIN 实现同样的效果 SELECT T1.continent, SUM(T2.population) FROM world AS T1 INNER JOIN world AS T2 ON T1.name = T2.name WHERE T2.population > 55000000 GROUP BY T1.continent; ``` ##### 示例 2: 列出欧洲各国首都名称与所在国名 这个问题适合采用左连接的方式来进行求解,即使有些国家可能暂时还没有定义其首都城市也不会丢失该国的信息。 ```sql SELECT country.name, city.name FROM country LEFT JOIN city ON country.capital = city.id WHERE country.continent='Europe'; ```
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