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MySQL：SQLZOO练习题解答

2020-08-17 10:56 951 查看

基本学习了SQL的查询语句，通过SQLZOO的题库做一个检查，下面记录了相关主题的问题与答案。

第一部分: SELECT basics

地址：https://sqlzoo.net/wiki/SELECT_basics

The example uses a WHERE clause to show the population of ‘France’. Note that strings (pieces of text that are data) should be in ‘single quotes’;
Modify it to show the population of Germany

SELECT population FROM world
WHERE name = 'Germany';

Checking a list The word IN allows us to check if an item is in a list. The example shows the name and population for the countries ‘Brazil’, ‘Russia’, ‘India’ and ‘China’.
Show the name and the population for ‘Sweden’, ‘Norway’ and ‘Denmark’.

SELECT name, population FROM world
WHERE name IN ('Sweden','Norway','Denmark');

Which countries are not too small and not too big? BETWEEN allows range checking (range specified is inclusive of boundary values). The example below shows countries with an area of 250,000-300,000 sq. km. Modify it to show the country and the area for countries with an area between 200,000 and 250,000.

SELECT name, area FROM world
WHERE area BETWEEN 200000 and 250000;

第二部分: SELECT from WORLD Tutorial

地址：https://sqlzoo.net/wiki/SELECT_from_WORLD_Tutorial

Read the notes about this table. Observe the result of running this SQL command to show the name, continent and population of all countries.

SELECT name, continent, population FROM world;

How to use WHERE to filter records. Show the name for the countries that have a population of at least 200 million. 200 million is 200000000, there are eight zeros.

SELECT name FROM world
WHERE population >= 200000000;

Give the name and the per capita GDP for those countries with a population of at least 200 million.
HELP:How to calculate per capita GDP
per capita GDP is the GDP divided by the population GDP/population

SELECT name, GDP/population AS per_GDP
FROM world WHERE population >= 200000000;

Show the name and population in millions for the countries of the continent ‘South America’. Divide the population by 1000000 to get population in millions.

SELECT name, population/1000000 AS popu_mil
FROM world WHERE continent  = 'South America';

Show the name and population for France, Germany, Italy.

SELECT name, population FROM world
WHERE name in ('fRANCE','Germany','Italy');

Show the countries which have a name that includes the word ‘United’

SELECT name FROM world
WHERE name LIKE '%United%';

Two ways to be big: A country is big if it has an area of more than 3 million sq km or it has a population of more than 250 million.
Show the countries that are big by area or big by population. Show name, population and area.

SELECT name, population, area FROM world
WHERE area>3000000 OR population >250000000;

Exclusive OR (XOR). Show the countries that are big by area (more than 3 million) or big by population (more than 250 million) but not both. Show name, population and area.
Australia has a big area but a small population, it should be included.
Indonesia has a big population but a small area, it should be included.
China has a big population and big area, it should be excluded.
United Kingdom has a small population and a small area, it should be excluded.

SELECT name, population, area FROM world
WHERE area > 3000000 XOR population > 250000000;

Show the name and population in millions and the GDP in billions for the countries of the continent ‘South America’. Use the ROUND function to show the values to two decimal places.
For South America show population in millions and GDP in billions both to 2 decimal places.
Millions and billions
Divide by 1000000 (6 zeros) for millions. Divide by 1000000000 (9 zeros) for billions.

SELECT name, ROUND(population/1000000,2) AS POP_mil, ROUND(GDP/1000000000,2) AS GDP_bil
FROM world  WHERE continent = 'South America';

Show the name and per-capita GDP for those countries with a GDP of at least one trillion (1000000000000; that is 12 zeros). Round this value to the nearest 1000.
Show per-capita GDP for the trillion dollar countries to the nearest $1000.

SELECT name, ROUND(GDP/population, -3) AS per_capita_GDP
FROM world
WHERE GDP>=1000000000000;

Greece has capital Athens.
Each of the strings ‘Greece’, and ‘Athens’ has 6 characters.
Show the name and capital where the name and the capital have the same number of characters.
You can use the LENGTH function to find the number of characters in a string

SELECT name, capital FROM world
WHERE length(name)=length(capital);

The capital of Sweden is Stockholm. Both words start with the letter ‘S’.
Show the name and the capital where the first letters of each match. Don’t include countries where the name and the capital are the same word.
You can use the function LEFT to isolate the first character.
You can use <> as the NOT EQUALS operator.

SELECT name, capital FROM world
WHERE LEFT(name,1) = LEFT(capital,1) AND name <> capital;

Equatorial Guinea and Dominican Republic have all of the vowels (a e i o u) in the name. They don’t count because they have more than one word in the name.
Find the country that has all the vowels and no spaces in its name.
You can use the phrase name NOT LIKE ‘%a%’ to exclude characters from your results.
The query shown misses countries like Bahamas and Belarus because they contain at least one ‘a’

SELECT name FROM world
WHERE name LIKE '%a%'
AND name LIKE '%e%'
AND name LIKE '%i%'
AND name LIKE '%o%'
AND name LIKE '%u%'
AND name NOT LIKE '% %';

第三部分: SELECT from Nobel Tutorial

地址：https://sqlzoo.net/wiki/SELECT_from_Nobel_Tutorial

Change the query shown so that it displays Nobel prizes for 1950.

SELECT yr, subject, winner
FROM nobel  WHERE yr = 1950;

Show who won the 1962 prize for Literature.

SELECT winner FROM nobel
WHERE yr = 1962 and subject= 'Literature';

Show the year and subject that won ‘Albert Einstein’ his prize.

SELECT yr, subject FROM nobel
WHERE winner = 'Albert Einstein';

Give the name of the ‘Peace’ winners since the year 2000, including 2000.

SELECT winner FROM nobel
WHERE subject = 'Peace' AND yr >=2000;

Show all details (yr, subject, winner) of the Literature prize winners for 1980 to 1989 inclusive.

SELECT * FROM nobel
WHERE subject = 'Literature' AND yr BETWEEN 1980 AND 1989;

Show all details of the presidential winners:
Theodore Roosevelt
Woodrow Wilson
Jimmy Carter
Barack Obama

SELECT * FROM nobel
WHERE winner IN (
'Theodore Roosevelt',
'Woodrow Wilson',
'Jimmy Carter',
'Barack Obama');

Show the winners with first name John

SELECT winner FROM nobel WHERE winner LIKE 'John%';

Show the year, subject, and name of Physics winners for 1980 together with the Chemistry winners for 1984.

SELECT yr, subject, winner FROM nobel
WHERE  (subject = 'Physics' AND yr= 1980)
OR (subject = 'Chemistry' AND yr=1984);

Show the year, subject, and name of winners for 1980 excluding Chemistry and Medicine

SELECT yr, subject, winner FROM nobel
WHERE yr=1980 and subject NOT IN ('Chemistry','Medicine');

Show year, subject, and name of people who won a ‘Medicine’ prize in an early year (before 1910, not including 1910) together with winners of a ‘Literature’ prize in a later year (after 2004, including 2004)

SELECT yr, subject, winner FROM nobel
WHERE (subject = 'Medicine' AND yr < 1910 )
OR (subject ='Literature' and yr >= 2004);

Find all details of the prize won by PETER GRÜNBERG
Non-ASCII characters

SELECT * FROM nobel
WHERE winner  = 'PETER GRÜNBERG';

Find all details of the prize won by EUGENE O’NEILL

SELECT * FROM nobel
WHERE winner = 'EUGENE O\'NEILL';

Knights of the realm
List the winners, year and subject where the winner starts with Sir. Show the the most recent first, then by name order.

SELECT winner, yr, subject FROM nobel
WHERE winner LIKE 'Sir%'
ORDER BY yr DESC, winner;

Chemistry and Physics last
The expression subject IN (‘Chemistry’,‘Physics’) can be used as a value - it will be 0 or 1.
Show the 1984 winners and subject ordered by subject and winner name; but list Chemistry and Physics last.

SELECT winner, subject FROM nobel
WHERE yr=1984
ORDER BY subject IN ('Physics','Chemistry'),subject,winner;

same to the result as below:

SELECT winner, subject FROM nobel
WHERE yr=1984
ORDER BY
CASE WHEN subject IN ('Physics','Chemistry') THEN 1
ELSE 0 END,
subject,winner;

第四部分: SELECT names

地址： https://napier.sqlzoo.net/wiki/SELECT_names

You can use WHERE name LIKE ‘B%’ to find the countries that start with “B”.
The % is a wild-card it can match any characters
Find the country that start with Y

SELECT name FROM world
WHERE name LIKE 'Y%';

Find the countries that end with y

SELECT name FROM world
WHERE name LIKE '%y';

3.Luxembourg has an x - so does one other country. List them both.
Find the countries that contain the letter x

SELECT name FROM world
WHERE name LIKE '%x%';

Iceland, Switzerland end with land - but are there others?
Find the countries that end with land

SELECT name FROM world
WHERE name LIKE '%land';

5.Columbia starts with a C and ends with ia - there are two more like this.
Find the countries that start with C and end with ia

SELECT name FROM world
WHERE name LIKE 'C%ia%';

Greece has a double e - who has a double o?
Find the country that has oo in the name

SELECT name FROM world
WHERE name LIKE '%oo%';

Bahamas has three a - who else?
Find the countries that have three or more a in the name

SELECT name FROM world
WHERE name LIKE '%a%a%a%';

India and Angola have an n as the second character. You can use the underscore as a single character wildcard.
Find the countries that have “t” as the second character.

SELECT name FROM world
WHERE name LIKE '_t%';

Lesotho and Moldova both have two o characters separated by two other characters.
Find the countries that have two “o” characters separated by two others.

SELECT name FROM world
WHERE name LIKE '%o__o%';

Cuba and Togo have four characters names.
Find the countries that have exactly four characters.

SELECT name FROM world
WHERE name LIKE '____';

The capital of Luxembourg is Luxembourg. Show all the countries where the capital is the same as the name of the country
Find the country where the name is the capital city.

SELECT name FROM world
WHERE name = capital;

The capital of Mexico is Mexico City. Show all the countries where the capital has the country together with the word “City”.
Find the country where the capital is the country plus “City”.

SELECT name FROM world
WHERE capital = CONCAT(name,' City');

Find the capital and the name where the capital includes the name of the country.

SELECT capital, name FROM world
WHERE capital LIKE CONCAT('%',name,'%');

Find the capital and the name where the capital is an extension of name of the country.
You should include Mexico City as it is longer than Mexico. You should not include Luxembourg as the capital is the same as the country.

SELECT capital, name FROM world
WHERE capital LIKE CONCAT(name,'%') AND capital != name;

For Monaco-Ville the name is Monaco and the extension is -Ville.
Show the name and the extension where the capital is an extension of name of the country.
You can use the SQL function REPLACE.

SELECT name,
REPLACE(capital,name,'') AS extension
FROM world
WHERE capital LIKE CONCAT(name,'%') AND capital != name;

第五部分: SELECT within SELECT Tutorial

地址：https://napier.sqlzoo.net/wiki/SELECT_within_SELECT_Tutorial

List each country name where the population is larger than that of ‘Russia’.
world(name, continent, area, population, gdp)

SELECT name FROM world
WHERE population >
(SELECT population FROM world
WHERE name='Russia');

Show the countries in Europe with a per capita GDP greater than ‘United Kingdom’.

SELECT name FROM world
WHERE continent = 'Europe' 	AND GDP/population >
(SELECT GDP/population FROM world
WHERE name = 'United Kingdom');

Neighbours of Argentina and Australia
List the name and continent of countries in the continents containing either Argentina or Australia. Order by name of the country.

SELECT name, continent FROM world
WHERE continent IN (
SELECT continent FROM world
WHERE name IN ('Argentina','Australia'))
ORDER BY name;

Between Canada and Poland
Which country has a population that is more than Canada but less than Poland? Show the name and the population.

SELECT name, population FROM world
WHERE population > (
SELECT population FROM world WHERE name = 'Canada')
AND population <
(SELECT population FROM world WHERE name = 'Poland');

Percentages of Germany
Germany (population 80 million) has the largest population of the countries in Europe. Austria (population 8.5 million) has 11% of the population of Germany.
Show the name and the population of each country in Europe. Show the population as a percentage of the population of Germany.

SELECT name,
CONCAT(
ROUND(100*population/(SELECT population FROM world WHERE name = 'Germany'),0),
'%') AS percentage
FROM world WHERE continent = 'Europe';

Bigger than every country in Europe
Which countries have a GDP greater than every country in Europe? [Give the name only.] (Some countries may have NULL gdp values)

SELECT name FROM world
WHERE GDP >
(SELECT MAX(GDP) FROM world WHERE continent = 'Europe' AND GDP>0);

Largest in each continent
Find the largest country (by area) in each continent, show the continent, the name and the area

SELECT continent, name, area FROM world x
WHERE area >= ALL
(SELECT area FROM world y
WHERE y.continent=x.continent
AND area>0);

8.First country of each continent (alphabetically)
List ach continent and the name of the country that comes first alphabetically.

SELECT w1.continent, w1.name FROM world w1
WHERE w1.name <= ALL
(SELECT w2.name FROM world w2
WHERE w1.continent = w2.continent)

Find the continents where all countries have a population <= 25000000. Then find the names of the countries associated with these continents. Show name, continent and population.

SELECT w1.name,w1.continent,w1.population
FROM world w1
WHERE 25000000>=
(SELECT MAX(w2.population) FROM world w2
WHERE w1.continent = w2.continent GROUP BY w2.continent)

Some countries have populations more than three times that of any of their neighbours (in the same continent). Give the countries and continents.

SELECT x.name, x.continent FROM world x
WHERE x.population >= ALL
(SELECT 3*y.population FROM world y
WHERE x.continent = y.continent AND x.name != y.name)

第六部分: SUM and COUNT

地址： https://napier.sqlzoo.net/wiki/SUM_and_COUNT

Total world population
Show the total population of the world.

SELECT SUM(population) FROM world

List of continents
List all the continents - just once each.

SELECT DISTINCT(continent) FROM world

GDP of Africa
Give the total GDP of Africa

SELECT SUM(GDP) FROM world WHERE continent = 'Africa';

Count the big countries
How many countries have an area of at least 1000000

SELECT COUNT(DISTINCT name) FROM world
WHERE area >= 1000000;

Baltic states population
What is the total population of (‘Estonia’, ‘Latvia’, ‘Lithuania’)

SELECT SUM(population) FROM worldWHERE name IN ('Estonia','Latvia','Lithuania');

Counting the countries of each continent
For each continent show the continent and number of countries.

SELECT continent, COUNT(name) AS num_country
FROM world GROUP BY continent;

Counting big countries in each continent
For each continent show the continent and number of countries with populations of at least 10 million.

SELECT continent, COUNT(name) FROM world
WHERE population >= 10000000
GROUP BY continent;

Counting big continents
List the continents that have a total population of at least 100 million.

SELECT continent FROM world
GROUP BY continent
HAVING SUM(population) >= 100000000;

第七部分: SUM and COUNT - nobel table

地址：https://napier.sqlzoo.net/wiki/The_nobel_table_can_be_used_to_practice_more_SUM_and_COUNT_functions.

Show the total number of prizes awarded.

SELECT COUNT(*) FROM nobel;

List each subject - just once

SELECT DISTINCT subject FROM nobel;

Show the total number of prizes awarded for Physics.

SELECT COUNT(*) FROM nobel
WHERE subject  = 'Physics';

For each subject show the subject and the number of prizes.

SELECT subject, COUNT(*) FROM nobel
GROUP BY subject;

For each subject show the first year that the prize was awarded.

SELECT subject, MIN(yr) FROM nobel
GROUP BY subject;

For each subject show the number of prizes awarded in the year 2000.

SELECT subject, COUNT(*) FROM nobel
WHERE yr= 2000
GROUP BY subject;

Show the number of different winners for each subject.

SELECT subject, COUNT(DISTINCT winner )
FROM nobel
GROUP BY subject;

For each subject show how many years have had prizes awarded.

SELECT subject, COUNT(DISTINCT yr)
FROM nobel
GROUP BY subject;

Show the years in which three prizes were given for Physics.

SELECT yr FROM nobel
WHERE subject = 'Physics'
GROUP BY yr
HAVING COUNT(subject) = 3;

Show winners who have won more than once.

SELECT winner  FROM nobel
GROUP BY winner
HAVING COUNT(subject) > 1;

Show winners who have won more than one subject.

SELECT winner FROM nobel
GROUP BY winner
HAVING COUNT(DISTINCT subject) > 1;

Show the year and subject where 3 prizes were given. Show only years 2000 onwards.

SELECT yr, subject  FROM nobel
WHERE  yr >=2000
GROUP BY yr, subject
HAVING COUNT(DISTINCT winner) = 3;

第七部分: The JOIN operation

地址：https://napier.sqlzoo.net/wiki/The_JOIN_operation

The first example shows the goal scored by a player with the last name ‘Bender’. The * says to list all the columns in the table - a shorter way of saying matchid, teamid, player, gtime
Modify it to show the matchid and player name for all goals scored by Germany. To identify German players, check for:
teamid = ‘GER’

SELECT matchid,player FROM goal
WHERE teamid  = 'GER';

From the previous query you can see that Lars Bender’s scored a goal in game 1012. Now we want to know what teams were playing in that match.Notice in the that the column matchid in the goal table corresponds to the id column in the game table. We can look up information about game 1012 by finding that row in the game table.Show id, stadium, team1, team2 for just game 1012

SELECT id,stadium,team1,team2
FROM game WHERE id = 1012

The code below shows the player (from the goal) and stadium name (from the game table) for every goal scored.Modify it to show the player, teamid, stadium and mdate for every German goal.

SELECT player,teamid,stadium, mdate
FROM game JOIN goal ON (id=matchid)
WHERE goal.teamid = 'GER';

Use the same JOIN as in the previous question.
Show the team1, team2 and player for every goal scored by a player called Mario player LIKE ‘Mario%’

SELECT team1, team2, player
FROM game ga
JOIN goal go ON ga.id = go.matchid
WHERE go.player LIKE 'Mario%';

The table eteam gives details of every national team including the coach. You can JOIN goal to eteam using the phrase goal JOIN eteam on teamid=id
Show player, teamid, coach, gtime for all goals scored in the first 10 minutes gtime<=10

SELECT player, teamid, coach, gtime
FROM goal go
JOIN eteam et ON go.teamid = et.id
WHERE gtime <= 10;

To JOIN game with eteam you could use either
game JOIN eteam ON (team1=eteam.id) or game JOIN eteam ON (team2=eteam.id)
Notice that because id is a column name in both game and eteam you must specify eteam.id instead of just id
List the dates of the matches and the name of the team in which ‘Fernando Santos’ was the team1 coach.

SELECT mdate, teamname
FROM game ga
JOIN eteam et ON ga.team1 = et.id
WHERE coach = 'Fernando Santos';

List the player for every goal scored in a game where the stadium was ‘National Stadium, Warsaw’

SELECT player FROM goal
JOIN game ON matchid = id
WHERE stadium = 'National Stadium, Warsaw';

The example query shows all goals scored in the Germany-Greece quarterfinal.Instead show the name of all players who scored a goal against Germany.

SELECT DISTINCT player
FROM game JOIN goal ON matchid = id
WHERE teamid != 'GER' AND (team1 = 'GER' OR team2 = 'GER');

Show teamname and the total number of goals scored.COUNT and GROUP BY

SELECT teamname, COUNT(matchid)
FROM eteam
JOIN goal ON id=teamid
GROUP BY teamname;

Show the stadium and the number of goals scored in each stadium.

SELECT stadium, COUNT(matchid)
FROM game
JOIN goal on id = matchid
GROUP BY stadium;

For every match involving ‘POL’, show the matchid, date and the number of goals scored.

SELECT matchid, mdate, COUNT(matchid)
FROM game
JOIN goal ON matchid = id
WHERE (team1 = 'POL' OR team2 = 'POL')
GROUP BY matchid, mdate;

For every match where ‘GER’ scored, show matchid, match date and the number of goals scored by ‘GER’

SELECT matchid, mdate, COUNT(matchid)
FROM goal
JOIN game ON id= matchid
WHERE teamid = 'GER'
GROUP BY matchid,mdate;

List every match with the goals scored by each team as shown. This will use “CASE WHEN” which has not been explained in any previous exercises.
Notice in the query given every goal is listed. If it was a team1 goal then a 1 appears in score1, otherwise there is a 0. You could SUM this column to get a count of the goals scored by team1. Sort your result by mdate, matchid, team1 and team2.

SELECT mdate,
team1,
SUM(CASE WHEN team1 = teamid THEN 1 ELSE 0 END) AS score1,
team2,
SUM(CASE WHEN team2 = teamid THEN 1 ELSE 0 END) AS score2
FROM game a
LEFT JOIN goal b ON a.id = b.matchid
GROUP BY mdate, team1,team2
ORDER BY mdate, matchid, team1, team2;

第八部分: Old JOIN Tutorial

地址：https://napier.sqlzoo.net/wiki/Old_JOIN_Tutorial

Show the athelete (who) and the country name for medal winners in 2000.

SELECT who, country.name
FROM ttms JOIN country ON (ttms.country=country.id)
WHERE games = 2000

Show the who and the color of the medal for the medal winners from ‘Sweden’.

SELECT who, color
FROM ttms t
JOIN country c ON t.country = c.id
WHERE c.name = 'Sweden';

Show the years in which ‘China’ won a ‘gold’ medal.

SELECT games
FROM ttms t
JOIN country c ON t.country = c.id
WHERE c.name = 'China' AND color = 'gold';

Show who won medals in the ‘Barcelona’ games.

SELECT who FROM ttws
JOIN games ON (ttws.games=games.yr)
WHERE games.city = 'Barcelona';

Show which city ‘Jing Chen’ won medals. Show the city and the medal color.

SELECT city,color
FROM games g
JOIN ttws t ON t.games = g.yr
WHERE t.who = 'Jing Chen';

Show who won the gold medal and the city.

SELECT who,city
FROM ttws t
JOIN games g ON g.yr = t.games
WHERE color='gold';

Show the games and color of the medal won by the team that includes ‘Yan Sen’.

SELECT games, color
FROM ttmd a
JOIN team b ON a.team = b.id
WHERE b.name LIKE '%Yan Sen%';

Show the ‘gold’ medal winners in 2004.

SELECT name
FROM team a
JOIN ttmd b ON a.id = b.team
WHERE games = 2004 AND color = 'gold';

Show the name of each medal winner country ‘FRA’.

SELECT name
FROM team a
JOIN ttmd b ON a.id = b.team
WHERE country = 'FRA';

第九部分: JOIN operation - Music Tutorial

地址：https://sqlzoo.net/wiki/Music_Tutorial

Find the title and artist who recorded the song ‘Alison’.

SELECT title, artist
FROM album
JOIN track ON (album.asin=track.album)
WHERE song = 'Alison'

Which artist recorded the song ‘Exodus’?

SELECT artist
FROM album
JOIN track ON asin = album
WHERE song = 'Exodus';

Show the song for each track on the album ‘Blur’

SELECT song
FROM track
JOIN album ON asin = album
WHERE title = 'Blur';

For each album show the title and the total number of track.

SELECT title, COUNT(*)
FROM album JOIN track ON (asin=album)
GROUP BY title

For each album show the title and the total number of tracks containing the word ‘Heart’ (albums with no such tracks need not be shown).

SELECT title, COUNT(*)
FROM album JOIN track ON album = asin
WHERE song LIKE '%Heart%'
GROUP BY title;

A “title track” is where the song is the same as the title. Find the title tracks.

SELECT song
FROM track JOIN album ON asin = album
WHERE song = title;

An “eponymous” album is one where the title is the same as the artist (for example the album ‘Blur’ by the band ‘Blur’). Show the eponymous albums.

SELECT title FROM album WHERE title = artist

Find the songs that appear on more than 2 albums. Include a count of the number of times each shows up.

SELECT song, COUNT(DISTINCT asin)
FROM track JOIN album ON asin = album
GROUP BY song
HAVING COUNT( DISTINCT asin) >2;

A “good value” album is one where the price per track is less than 50 pence. Find the good value album - show the title, the price and the number of tracks.

SELECT title, price, COUNT(album)
FROM album JOIN track ON asin = album
GROUP BY title, price
HAVING price/COUNT(album) < 0.5

Wagner’s Ring cycle has an imposing 173 tracks, Bing Crosby
clocks up 101 tracks.
List albums so that the album with the most tracks is first.
Show the title and the number of tracksWhere two or more albums have the same number of tracks you should order alphabetically

SELECT title, COUNT(album)
FROM album JOIN track ON asin = album
GROUP BY title
ORDER BY COUNT(album) DESC, title;

第十部分: More JOIN operations - Movie table

地址：https://sqlzoo.net/wiki/More_JOIN_operations

List the films where the yr is 1962 [Show id, title]

SELECT id, title   FROM movie  WHERE yr=1962

Give year of ‘Citizen Kane’.

SELECT yr FROM movie WHERE title = 'Citizen Kane';

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;

What id number does the actor ‘Glenn Close’ have?

SELECT id FROM actor WHERE name = 'Glenn Close';

What is the id of the film ‘Casablanca’

SELECT id FROM movie WHERE title = 'Casablanca';

Obtain the cast list for ‘Casablanca’.
what is a cast list?Use movieid=11768, (or whatever value you got from the previous question)

SELECT a.name
FROM actor a
JOIN casting c ON a.id = c.actorid
JOIN movie m ON m.id = c.movieid
WHERE m.title = 'Casablanca';

Obtain the cast list for the film ‘Alien’

SELECT a.name
FROM actor a
JOIN casting c ON a.id = c.actorid
JOIN movie m ON m.id = c.movieid
WHERE m.title = 'Alien';

List the films in which ‘Harrison Ford’ has appeared

SELECT title
FROM movie m
JOIN casting c ON c.movieid = m.id
JOIN actor a ON a.id = c.actorid
WHERE a.name = 'Harrison Ford';

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 c.movieid = m.id
JOIN actor a ON a.id = c.actorid
WHERE a.name = 'Harrison Ford' AND c.ord != 1;

List the films together with the leading star for all 1962 films.

SELECT m.title, a.name
FROM movie m
JOIN casting c ON c.movieid = m.id
JOIN actor a ON a.id = c.actorid
WHERE c.ord = 1 AND yr = 1962;

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(m.id)
FROM movie m
JOIN casting c ON m.id = c.movieid
JOIN actor a ON a.id = c.actorid
WHERE a.name = 'Rock Hudson'
GROUP BY yr
HAVING COUNT(m.id) >2;

List the film title and the leading actor for all of the films ‘Julie Andrews’ played in

SELECT title, a.name
FROM movie m
JOIN casting c ON m.id = c.movieid
JOIN actor a ON a.id = c.actorid
WHERE c.ord =1 AND m.id IN
(SELECT c.movieid
FROM casting c JOIN actor a ON a.id = c.actorid
WHERE a.name = 'Julie Andrews');

Obtain a list, in alphabetical order, of actors who’ve had at least 15 starring roles.

SELECT name FROM actor a
JOIN casting c ON a.id = c.actorid
WHERE c.ord = 1
GROUP BY name
HAVING COUNT(c.ord) >=15;

List the films released in the year 1978 ordered by the number of actors in the cast, then by title.

SELECT title, COUNT(c.actorid)
FROM movie m
JOIN casting c ON c.movieid = m.id
WHERE yr = 1978
GROUP BY title
ORDER BY COUNT(c.actorid) DESC, title;

List all the people who have worked with ‘Art Garfunkel’.

SELECT name
FROM actor a
JOIN casting c ON a.id = c.actorid
WHERE name != 'Art Garfunkel' AND c.movieid IN
(SELECT movieid
FROM casting c JOIN actor a ON a.id = c.actorid
WHERE a.name = 'Art Garfunkel');

第十一部分: Using Null

地址：https://sqlzoo.net/wiki/Using_Null

List the teachers who have NULL for their department.

SELECT name FROM teacher WHERE dept IS NULL;

Note the INNER JOIN misses the teachers with no department and the departments with no teacher.

SELECT teacher.name, dept.name
FROM teacher
INNER JOIN dept ON (teacher.dept=dept.id)

Use a different JOIN so that all teachers are listed.

SELECT t.name, d.name
FROM teacher t
LEFT JOIN dept d ON t.dept = d.id;

Use a different JOIN so that all departments are listed.

SELECT t.name, d.name
FROM teacher t
RIGHT JOIN dept d ON d.id = t.dept;

Using the COALESCE function
Use COALESCE to print the mobile number. Use the number ‘07986 444 2266’ if there is no number given.
Show teacher name and mobile number or ‘07986 444 2266’

SELECT name, COALESCE(mobile,'07986 444 2266') as mob
FROM teacher

Use the COALESCE function and a LEFT JOIN to print the teacher name and department name.
Use the string ‘None’ where there is no department.

SELECT t.name, COALESCE(d.name, 'None') as dept_name
FROM teacher t
LEFT JOIN dept d ON d.id = t.dept;

Use COUNT to show the number of teachers and the number of mobile phones.

SELECT COUNT(id) t_num, COUNT(mobile) m_num
FROM teacher;

Use COUNT and GROUP BY dept.name to show each department and the number of staff.
Use a RIGHT JOIN to ensure that the Engineering department is listed.

SELECT d.name, COUNT(t.id)
FROM teacher t
RIGHT JOIN dept d ON d.id = t.dept
GROUP BY d.name;

Use CASE to show the name of each teacher followed by ‘Sci’ if the teacher is in dept 1 or 2 and ‘Art’ otherwise.

SELECT name,
CASE WHEN dept = 1 OR dept = 2 THEN 'Sci'
ELSE 'Art' END
FROM teacher

Use CASE to show the name of each teacher followed by ‘Sci’ if the teacher is in dept 1 or 2, show ‘Art’ if the teacher’s dept is 3 and ‘None’ otherwise.

SELECT name,
CASE WHEN dept IN (1,2) THEN 'Sci'
WHEN dept IS NULL THEN 'None'
ELSE 'Art'
END
FROM teacher;

第十二部分: Self join

地址：https://sqlzoo.net/wiki/Self_join

How many stops are in the database.

SELECT COUNT(id) FROM stops;

Find the id value for the stop ‘Craiglockhart’

SELECT id FROM stops
WHERE name = 'Craiglockhart';

Give the id and the name for the stops on the ‘4’ ‘LRT’ service.

SELECT id, name
FROM stops s
JOIN route r ON s.id = r.stop
WHERE num = '4' AND company = 'LRT';

Routes and stops
The query shown gives the number of routes that visit either London Road (149) or Craiglockhart (53). Run the query and notice the two services that link these stops have a count of 2.
Add a HAVING clause to restrict the output to these two routes.

SELECT company, num, COUNT(*)
FROM route WHERE stop=149 OR stop=53
GROUP BY company, num
HAVING COUNT(*) = 2;

Execute the self join shown and observe that b.stop gives all the places you can get to from Craiglockhart, without changing routes.
Change the query so that it shows the services from Craiglockhart to London Road.

SELECT a.company, a.num, a.stop, b.stop
FROM route a
JOIN route b ON  (a.company=b.company AND a.num=b.num)
WHERE a.stop=53 AND b.stop = 149;

The query shown is similar to the previous one, however by joining two copies of the stops table we can refer to stops by name rather than by number.
Change the query so that the services between ‘Craiglockhart’ and ‘London Road’ are shown. If you are tired of these places try ‘Fairmilehead’ against ‘Tollcross’

SELECT a.company, a.num, stopa.name, stopb.name
FROM route a JOIN route b ON
(a.company=b.company AND a.num=b.num)
JOIN stops stopa ON (a.stop=stopa.id)
JOIN stops stopb ON (b.stop=stopb.id)
WHERE stopa.name='Craiglockhart' AND stopb.name = 'London Road';

Give a list of all the services which connect stops 115 and 137 (‘Haymarket’ and ‘Leith’)

SELECT DISTINCT a.company, a.num
FROM route a
JOIN route b ON a.company = b.company AND a.num = b.num
WHERE (a.stop = 115 AND b.stop = 137) OR (a.stop = 137 AND b.stop = 115);

Give a list of the services which connect the stops ‘Craiglockhart’ and ‘Tollcross’

SELECT DISTINCT r1.company, r1.num
FROM route r1
JOIN route r2 ON r1.company = r2.company AND r1.num = r2.num
JOIN stops s1 ON r1.stop = s1.id
JOIN stops s2 ON r2.stop = s2.id
WHERE (s1.name = 'Craiglockhart' AND s2.name = 'Tollcross') OR (s2.name = 'Craiglockhart' AND s1.name = 'Tollcross');

Give a distinct list of the stops which may be reached from ‘Craiglockhart’ by taking one bus, including ‘Craiglockhart’ itself, offered by the LRT company. Include the company and bus no. of the relevant services.

SELECT DISTINCT d.name, a.company, a.num
FROM route a
JOIN route b ON a.company = b.company AND a.num = b.num
JOIN stops c ON c.id = a.stop
JOIN stops d ON d.id = b.stop
WHERE a.company = 'LRT' AND (c.name ='Craiglockhart')

Find the routes involving two buses that can go from Craiglockhart to Lochend.
Show the bus no. and company for the first bus, the name of the stop for the transfer, and the bus no. and company for the second bus.
Hint
Self-join twice to find buses that visit Craiglockhart and Lochend, then join those on matching stops.

SELECT DISTINCT bus1.num, bus1.company, bus1.name, bus2.num, bus2.company
FROM
(SELECT r1.num, r1.company, s2.name, r2.stop
FROM route r1
JOIN route r2 ON r1.company = r2.company AND r1.num = r2.num
JOIN stops s1 ON s1.id = r1.stop
JOIN stops s2 ON s2.id = r2.stop
WHERE s1.name = 'Craiglockhart' AND s2.name != 'Lochend') AS bus1
JOIN
(SELECT r3.num, r3.stop, r4.company
FROM route r3
JOIN route r4 ON r3.num = r4.num AND r3.company = r4.company
JOIN stops s3 ON s3.id = r3.stop
JOIN stops s4 ON s4.id = r4.stop
WHERE s4.name = 'Lochend' AND s3.name != 'Craiglockhart') AS bus2
ON bus1.stop = bus2.stop
ORDER BY bus1.num,bus1.company,name,bus2.num;

--- 拆分为3个阶段：bus1 和 bus2，关联bus1和bus2
--- bus1： 从Crai站出发，到达任意一个不是Loch的站，汇总所有的路线，获取bus1的num,company,到达站(id,辅助)
--- bus2: 从任意一个不是Crai的站出发，达到Loch，汇总所有的路线, 获取bus2的num,company，出发站(id，辅助)
--- bus1 与 bus2关联：bus1的到达站 = bus2的出发站，bus1.stop(id) = bus2.stop (id)
--- 按照关键字排序

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