斯坦福大学sql练习（基础篇）

 这是斯坦福大学的公开课上关于sql数据库知识的基础测试，看看计算机顶尖学府是怎么出题的？真心有水平。

You've started a new movie-rating website, and you've been collecting data on reviewers' ratings of various movies. There's not much data yet, but you can still try out some interesting queries. Here's the schema:

Movie ( mID, title, year, director )
English: There is a movie with ID number mID, a title, a release year, and a director.

Reviewer ( rID, name )
English: The reviewer with ID number rID has a certain name.

Rating ( rID, mID, stars, ratingDate )
English: The reviewer rID gave the movie mID a number of stars rating (1-5) on a certain ratingDate.

 

Movie

mID	title	year	director
101	Gone with the Wind	1939	Victor Fleming
102	Star Wars	1977	George Lucas
103	The Sound of Music	1965	Robert Wise
104	E.T.	1982	Steven Spielberg
105	Titanic	1997	James Cameron
106	Snow White	1937	<null>
107	Avatar	2009	James Cameron
108	Raiders of the Lost Ark	1981	Steven Spielberg

Reviewer

rID	name
201	Sarah Martinez
202	Daniel Lewis
203	Brittany Harris
204	Mike Anderson
205	Chris Jackson
206	Elizabeth Thomas
207	James Cameron
208	Ashley White

Rating（这里的stars是几星的意思，就是评价）

rID	mID	stars	ratingDate
201	101	2	2011-01-22
201	101	4	2011-01-27
202	106	4	<null>
203	103	2	2011-01-20
203	108	4	2011-01-12
203	108	2	2011-01-30
204	101	3	2011-01-09
205	103	3	2011-01-27
205	104	2	2011-01-22
205	108	4	<null>
206	107	3	2011-01-15
206	106	5	2011-01-19
207	107	5	2011-01-20
208	104	3	2011-01-02

Question 1：Find the titles of all movies directed by Steven Spielberg.

select title from Movie
where director='Steven Spielberg

Question 2：Find all years that have a movie that received a rating of 4 or 5, and sort them in increasing order.

select distinct Movie.year from Movie
join Rating
on Movie.mID=Rating.mID and (Rating.stars=4 or Rating.stars=5)
order by Movie.year

Question 3：Find the titles of all movies that have no ratings.

select distinct Movie.title from Movie
left join Rating
on Movie.mID=Rating.mID
where Rating.mID is null

Question 4：Some reviewers didn't provide a date with their rating. Find the names of all reviewers who have ratings with a NULL value for the date.

select Reviewer.name from Reviewer
join Rating
on Reviewer.rID=Rating.rID
where Rating.ratingDate is null

Question 5：Write a query to return the ratings data in a more readable format: reviewer name, movie title, stars, and ratingDate. Also, sort the data, first by reviewer name, then by movie title, and lastly by number of stars.

select Reviewer.name,Movie.title,Rating.stars,Rating.ratingDate from 
(Reviewer join Rating) join Movie
on Reviewer.rID=Rating.rID and Rating.mID=Movie.mID
order by Reviewer.name,Movie.title,Rating.stars

Question 6:For all cases where the same reviewer rated the same movie twice and gave it a higher rating the second time, return the reviewer's name and the title of the movie.

select  Reviewer.name,Movie.title from Movie join
(select Rating.mID,Rating.rID,Rating.stars,Rating.ratingDate from Rating,
(select Rating.mID,Rating.rID,max(Rating.stars) as maxStars,max(Rating.ratingDate) as maxDate from Rating
group by mID,rID
having count(rId)=2 and count(mID)=2) as nan
where Rating.mID=nan.mID and Rating.rID=nan.rID and Rating.stars=nan.maxStars and Rating.ratingDate=maxDate) as nannan
on Movie.mID=nannan.mID join Reviewer on Reviewer.rID=nannan.rID
或者：
select Movie.title,Reviewer.name from Movie join
(select Rating.rID,Rating.mID from Rating,
(select Rating.rID,Rating.mID,max(Rating.stars) as maxStar,max(Rating.ratingDate) as maxDate from Rating
group by Rating.rID,Rating.mID
having count(rID)=2) as nan
where nan.maxStar=Rating.Stars and nan.maxDate=Rating.ratingDate) as nannan
on Movie.mID=nannan.mID
join Reviewer on Reviewer.rID=nannan.rID

Question 9:For each movie that has at least one rating, find the highest number of stars that movie received. Return the movie title and number of stars. Sort by movie title.

select Movie.title,nan.maxStars from Movie join
(select Rating.mID,max(Rating.stars) as maxStars from Rating
group by mID) as nan
on Movie.mID=nan.mID
ORDER BY Movie.title

Question 8:List movie titles and average ratings, from highest-rated to lowest-rated. If two or more movies have the same average rating, list them in alphabetical order.

select Movie.title,nan.avgStars from Movie join
(select Rating.mID,avg(Rating.stars) as avgStars from Rating
group by Rating.mID
) as nan
on Movie.mID=nan.mID
order by nan.avgStars desc,Movie.title

Question 9:Find the names of all reviewers who have contributed three or more ratings. (As an extra challenge, try writing the query without HAVING or without COUNT.)
常规方法：

select Reviewer.name from Reviewer join 
(select Rating.rID,count(Rating.rID) as countRID from Rating
group by rID
having countRID>=3) as nan
on nan.rID=Reviewer.rID

非常规方法暂时还没有想到。

Question 10:For each movie, return the title and the 'rating spread', that is, the difference between highest and lowest ratings given to that movie. Sort by rating spread from highest to lowest, then by movie title.

select Movie.title,nan.spread from Movie join
(select Rating.mID,(max(Rating.stars)-min(Rating.stars)) as spread from Rating
group by Rating.mID)as nan
on Movie.mID=nan.mID
order by nan.spread desc,Movie.title

未完待续。。。

SELECT tbl_info.name, tbl_info.num, tbl_info.phone, tbl_info.faculty, tbl_info.qq, tbl_grade.name, tbl_class.name
FROM tbl_info
JOIN tbl_grade
JOIN tbl_class ON tbl_info.grade_id = tbl_grade.id
AND tbl_info.class_id = tbl_class.id

。。。