SQL自学网习题答案

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附学习网址：自学SQL网(教程 视频 练习全套)

第一节：

1【初体验】这是第一题,请你先将左侧的输入框里的内容清空，然后请输入下面的SQL，您将看到所有电影标题：

答案：SELECT * FROM movies;

2【初体验】请输入如下SQL你将看到4条电影（切记先清空数据框且出错要耐心比对）：

答案：SELECT title,director FROM movies WHERE Id < 5

3【初体验】输入如下SQL你将看到电影总条数：

答案：SELECT count(*) FROM movies

4【初体验】SQL可以直接做计算，下面的SQL计算1+1的和,请输入：

答案：SELECT 1+1

 

第二节：

1【简单查询】找到所有电影的名称

答案：SELECT title FROM movies

2【简单查询】找到所有电影的导演

答案：SELECT Director FROM movies

3【简单查询】找到所有电影的名称和导演

答案：SELECT Director,title FROM movies

4【简单查询】找到所有电影的名称和上映年份

答案：SELECT title,Year FROM movies

5【简单查询】找到所有电影的所有信息

答案：SELECT * FROM movies

6【简单查询】找到所有电影的名称,Id和播放时长

答案：SELECT title,Id,Length_minutes FROM movies

 

第三节：

1【简单条件】找到id为6的电影

答案：SELECT * FROM movies where Id=6

2【简单条件】找到在2000-2010年间year上映的电影

答案：SELECT * FROM movies where Year>=2000 and Year<=2010

3【简单条件】找到不是在2000-2010年间year上映的电影

答案：SELECT * FROM movies where Year<2000 or Year >2010

4【简单条件】找到头5部电影

答案：SELECT * FROM movies where Id<=5

5【简单条件】找到2010（含）年之后的电影里片长小于两个小时的片子

答案：SELECT * FROM movies where Year>=2010 and Length_minutes<120

 

第四节：

1【复杂条件】找到所有Toy Story系列电影

答案：SELECT * FROM movies where Title LIKE “%Toy Story%”

2【复杂条件】找到所有John Lasseter导演的电影

答案：SELECT * FROM movies where Director =‘John Lasseter’

3【复杂条件】找到所有不是John Lasseter导演的电影

答案：SELECT * FROM movies where Director !=‘John Lasseter’

4【复杂条件】找到所有电影名为 “WALL-” 开头的电影

答案：SELECT * FROM movies where Title LIKE ‘WALL-%’

5【复杂条件】有一部98年电影中文名《虫虫危机》请给我找出来

答案：SELECT * FROM movies where Title = “A Bug’s Life”

 

第五节：

1【结果排序】按导演名排重列出所有电影(只显示导演)，并按导演名正序排列

答案：SELECT DISTINCT Director FROM movies ORDER BY Director ASC

2【结果排序】列出按上映年份最新上线的4部电影

答案：SELECT * FROM movies ORDER BY Year DESC limit 4

3【结果排序】按电影名字母序升序排列，列出前5部电影

答案：SELECT * FROM movies ORDER BY Title ASC limit 5

4【结果排序】按电影名字母序升序排列，列出上一题之后的5部电影

答案：SELECT * FROM movies ORDER BY Title ASC limit 5 OFFSET 5

5【结果排序】如果按片长排列，John Lasseter导演导过片长第3长的电影是哪部，列出名字即可

答案：SELECT Title FROM movies WHERE Director=‘John Lasseter’ ORDER BY Length_minutes ASC limit 1 offset 2

 

第六节：

1【复习】列出所有加拿大人的Canadian信息(包括所有字段)

答案：SELECT * FROM north_american_cities WHERE Country = ‘Canada’

2【复习】列出所有在Chicago西部的城市，从西到东排序(包括所有字段)

答案：SELECT * FROM north_american_cities WHERE Longitude<-87.629798 ORDER BY Longitude DESC

3【复习】用人口数population排序,列出墨西哥Mexico最大的2个城市(包括所有字段)

答案：SELECT * FROM North_american_cities WHERE Country=‘Mexico’ ORDER BY Population ASC LIMIT 2 OFFSET 1

4【复习】列出美国United States人口3-4位的两个城市和他们的人口(包括所有字段)

答案：SELECT * FROM North_american_cities WHERE country = ‘United States’ ORDER BY Population DESC LIMIT 2 OFFSET 2

 

第七节：

1【联表】找到所有电影的国内Domestic_sales和国际销售额

答案：SELECT * FROM movies inner join boxoffice on movies.id = boxoffice.movie_id

2【联表】找到所有国际销售额比国内销售大的电影

答案：SELECT * FROM movies inner join Boxoffice on Movies.Id = Boxoffice.Movie_id WHERE Domestic_sales<International_sales

3【联表】找出所有电影按市场占有率rating倒序排列 *****

答案：SELECT * FROM movies inner join Boxoffice on Movies.Id = Boxoffice.Movie_id ORDER BY rating DESC

4【联表】每部电影按国际销售额比较,排名最靠前的导演是谁,国际销量多少

答案：SELECT Director,International_sales FROM movies inner join Boxoffice on Movies.Id=Boxoffice.Movie_id ORDER BY International_sales DESC LIMIT 1

 

第八节： 

1【复习】找到所有有雇员的办公室(buildings)名字

答案：SELECT DISTINCT building FROM employees where building is not null

2【复习】找到所有办公室里的所有角色（包含没有雇员的）,并做唯一输出(DISTINCT)

答案：SELECT DISTINCT buildings.building_name, employees.Role FROM buildings left join employees on buildings.building_name = employees.building

3【难题】找到所有有雇员的办公室(buildings)和对应的容量

答案：SELECT DISTINCT Buildings.Building_name,Buildings.Capacity FROM Employees inner JOIN Buildings ON Employees.Building=Buildings.Building_name WHERE Building is not null

 

第九节：

1【复习】找到雇员里还没有分配办公室的(列出名字和角色就可以)

答案：SELECT Role,Name FROM employees WHERE Building is null

2【难题】找到还没有雇员的办公室

答案：SELECT Building_name FROM buildings left join employees on buildings.building_name = employees.building WHERE Role IS NULL

 

第十节：

1【计算】列出所有的电影ID,名字和销售总额(以百万美元为单位计算)

答案：SELECT Id,Title,(Domestic_sales+International_sales)/1000000 as 销售总额 FROM movies inner join Boxoffice on Movies.Id=Boxoffice.Movie_id

2【计算】列出所有的电影ID,名字和市场指数(Rating的10倍为市场指数)

答案：SELECT Id,Title,rating*10 as 市场指数 FROM movies inner join Boxoffice on Movies.Id=Boxoffice.Movie_id

3【计算】列出所有偶数年份的电影，需要电影ID,名字和年份

答案：SELECT Id,Title,year FROM movies inner join Boxoffice on Movies.Id=Boxoffice.Movie_id WHERE Year%2=0

4【难题】John Lasseter导演的每部电影每分钟值多少钱,告诉我最高的3个电影名和价值就可以

答案：SELECT Title,(Domestic_sales+International_sales)/Length_minutes as price FROM movies inner join Boxoffice on Movies.Id=Boxoffice.Movie_id WHERE Director=‘John Lasseter’ order by price DESC limit 3

 

第十一节：

1【统计】找出就职年份最高的雇员(列出雇员名字+年份）

答案：SELECT Name,max(Years_employed) FROM employees

2【分组】按角色(Role)统计一下每个角色的平均就职年份

答案：SELECT role,avg(Years_employed) FROM employees group by role

3【分组】按办公室名字总计一下就职年份总和

答案：SELECT Building,sum(Years_employed) FROM employees group by Building

4【难题】每栋办公室按人数排名,不要统计无办公室的雇员

答案：SELECT Building,count(Building)as count FROM employees group by Building order by count DESC limit 2

｛SELECT building,count(*) as count FROM employees where building is NOT null group by building

｝

 

第十二节：

1【统计】统计一下Artist角色的雇员数量

答案：SELECT count() FROM employees WHERE Role = ‘Artist’

2【分组】按角色统计一下每个角色的雇员数量

答案：SELECT role,count() FROM employees group by role

3【分组】算出Engineer角色的就职年份总计

答案：SELECT sum(Years_employed) FROM employees where role='Engineer’group by Role

4【难题】按角色分组算出每个角色按有办公室和没办公室的统计人数(列出角色，数量，有无办公室,注意一个角色如果部分有办公室，部分没有需分开统计）

答案：SELECT count(*) as count,Role,building is not null as bn FROM employees group by Role,bn

 

第十三节：

1【复习】统计出每一个导演的电影数量（列出导演名字和数量）

答案：SELECT DIRECTOr, count() FROM movies group by director

2【复习】统计一下每个导演的销售总额(列出导演名字和销售总额)

答案：SELECT DIRECTOr,sum(Domestic_sales)+sum(International_sales) as 销售总额 FROM movies inner join Boxoffice on Movies.Id = Boxoffice.Movie_id group by Director

3【难题】按导演分组计算销售总额,求出平均销售额冠军（统计结果过滤掉只有单部电影的导演，列出导演名，总销量，电影数量，平均销量)

答案：SELECT sum(Domestic_sales+International_sales) as sum_sale,director,count() as count, sum(Domestic_sales+International_sales)/count() as avg_sale FROM movies left join boxoffice on movies.id = boxoffice.movie_id group by director having count > 1 order by avg_sale desc limit 1

**** SELECT Director, SUM(Domestic_sales)+SUM(International_sales) AS TOTAL,count() AS COUNTS,(SUM(Domestic_sales)+SUM(International_sales))/count(*) as avg FROM MOVIES LEFT join Boxoffice on movies.Id=Boxoffice.movie_id group by director having countS > 1 order by avg DESC LIMIT 1

4【变态难】找出每部电影和单部电影销售冠军之间的销售差，列出电影名，销售额差额

答案：SELECT (SELECT (Domestic_sales+International_sales) as total_sale FROM movies left join boxoffice on movies.id = boxoffice.movie_id order by total_sale desc limit 1) - (Domestic_sales+International_sales) as sale_diff,title FROM movies left join boxoffice on movies.id = boxoffice.movie_id order by sale_diff desc

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标签：电影,sales,Boxoffice,movies,答案,SQL,习题,自学,SELECT
来源： https://www.cnblogs.com/txq521/p/16592518.html