SELECT from Nobel

1. 更改查詢以顯示1950年諾貝爾獎的獎項資料。
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  SELECT *
  FROM nobel
  where yr=1950

顯示誰贏得了1962年文學獎(Literature)。

SELECT winner
  FROM nobel
 WHERE yr = 1962
   AND subject = 'Literature'

1. 顯示“愛因斯坦”(‘Albert Einstein’) 的獲獎年份和獎項。
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  SELECT yr, subject
  FROM nobel
  WHERE winner='Albert Einstein'

顯示2000年及以後的和平獎(‘Peace’)得獎者。

SELECT winner
  FROM nobel
 WHERE yr>1999
  AND subject='Peace'

1. 顯示1980年至1989年(包含首尾)的文學獎(Literature)獲獎者所有細節（年，主題，獲獎者）。
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  select *
  from nobel
  where yr >=1980 and yr <= 1989
  and subject='Literature'

顯示總統獲勝者的所有細節：

西奧多•羅斯福 Theodore Roosevelt
伍德羅•威爾遜 Woodrow Wilson

吉米•卡特 Jimmy Carter

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

1. 顯示名字為John 的得獎者。 (注意:外國人名字(First name)在前，姓氏(Last name)在後)
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  select winner
  from nobel
  where winner like 'John%'

顯示1980年物理學(physics)獲獎者，及1984年化學獎(chemistry)獲得者。

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select *
from nobel
where (yr=1980 and subject='physics') or (yr=1984 and subject='chemistry')

查看1980年獲獎者，但不包括化學獎(Chemistry)和醫學獎(Medicine)。

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select *
from nobel
where subject not in ('chemistry', 'medicine') and yr=1980

1. 顯示早期的醫學獎(Medicine)得獎者（1910之前，不包括1910），及近年文學獎(Literature)得獎者（2004年以後，包括2004年）。
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  select *
  from nobel
  where (yr<1910 and subject='Medicine')
  or (yr >= 2004 and subject='Literature')

Find all details of the prize won by PETER GRÜNBERG

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select *
from nobel
where winner='PETER GRÜNBERG'

1. 查找尤金•奧尼爾EUGENE O’NEILL得獎的所有細節 Find all details of the prize won by EUGENE O’NEILL
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  select *
  from nobel
  where winner='EUGENE O\'NEILL'
1. 騎士列隊 Knights in order
  - 列出爵士的獲獎者、年份、獎頁(爵士的名字以Sir開始)。先顯示最新獲獎者，然後同年再按名稱順序排列。
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    select winner, yr, subject
    from nobel
    where winner like 'Sir%'
    order by yr DESC, winner
1. 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.
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    SELECT winner, subject
    FROM nobel
    WHERE yr=1984
    ORDER BY subject in ('Physics', 'Chemistry'), subject, winner

SELECT IN SELECT

List each country name where the population is larger than that of ‘Russia’.

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')

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

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')

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.

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

1. Which countries have a GDP greater than every country in Europe? [Give the name only.] (Some countries may have NULL gdp values)
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  SELECT name
  FROM world
  WHERE gdp > (
  SELECT MAX(gdp) FROM world WHERE continent='Europe'
  AND gdp IS NOT NULL)

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)

List each continent and the name of the country that comes first alphabetically.

SELECT continent, name
FROM world x
WHERE name <= ALL (
SELECT name FROM world y WHERE x.continent=y.continent)

或者

SELECT continent, name
FROM world x
WHERE x.name = (
SELECT name FROM world y WHERE x.continent=y.continent 
ORDER BY name 
LIMIT 1
)

1. 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.
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  SELECT name, continent, population
  FROM world x
  WHERE 25000000 >= ALL (
  SELECT y.population FROM world y WHERE x.continent=y.continent
  AND y.population >=0
  )

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

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

SUM and COUNT

This tutorial is about aggregate functions such as COUNT, SUM and AVG. An aggregate function takes many values and delivers just one value. For example the function SUM would aggregate the values 2, 4 and 5 to deliver the single value 11.

1. Show the total population of the world.
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  SELECT SUM(population)
  FROM world
1. List all the continents - just once each.
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  SELECT DISTINCT continent
  FROM world

Give the total GDP of Africa

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SELECT SUM(gdp)
FROM world
WHERE continent='Africa'

How many countries have an area of at least 1000000

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

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

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SELECT SUM(population)
FROM world
WHERE name in ('Estonia', 'Latvia', 'Lithuania')

1. For each continent show the continent and number of countries.
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  SELECT continent, COUNT(name)
  FROM world
  GROUP BY continent
1. For each continent show the continent and number of countries with populations of at least 10 million.
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  SELECT continent, COUNT(name)
  FROM world
  WHERE population >= 10000000
  GROUP BY continent

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

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

SUM and COUNT Quiz

选择题答案：

Q\A	1	2	3	4	5
1			√
2	√
3				√
4					√
5		√
6					√
7				√
8				√

JOIN

1. 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’

1 2	SELECT matchid, player FROM goal WHERE teamid='GER'

1. 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

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

3.You can combine the two steps into a single query with a JOIN.
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SELECT *
FROM game JOIN goal ON (id=matchid)

The FROM clause says to merge data from the goal table with that from the game table. The ON says how to figure out which rows in game go with which rows in goal - the id from goal must match matchid from game. (If we wanted to be more clear/specific we could say
ON (game.id=goal.matchid)

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.

1 2	SELECT player, teamid, stadium, mdate FROM game JOIN goal ON (game.id=goal.matchid and goal.teamid='GER')

1. 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%’

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SELECT team1, team2, player
FROM game JOIN goal
ON (id=matchid AND player LIKE 'Mario%')

1. 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

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SELECT player, teamid, coach, gtime
FROM goal JOIN eteam
ON (teamid=id AND gtime<=10)

1. 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 the dates of the matches and the name of the team in which ‘Fernando Santos’ was the team1 coach.

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SELECT mdate, teamname
FROM game JOIN eteam
ON (team1=eteam.id AND coach='Fernando Santos')

1. List the player for every goal scored in a game where the stadium was ‘National Stadium, Warsaw’
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  SELECT player
  FROM goal JOIN game
  ON (matchid = id AND stadium='National Stadium, Warsaw')
1. 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.
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  SELECT DISTINCT player
  FROM goal JOIN game
  ON matchid=id AND teamid <> 'GER' AND (team1='GER' OR team2='GER')

Show teamname and the total number of goals scored.

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SELECT teamname, COUNT(gtime)
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(*) as goals
FROM game JOIN goal
ON id=matchid AND (team1=teamid OR team2=teamid)
GROUP BY stadium

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

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

1. 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.

mdate	team1	score1	team2	score2
1	July	2012	ESP 4	ITA 0
10	June	2012	ESP 1	ITA 1
10	June	2012	IRL 1	CRO 3

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

JOIN_Quiz

选择题答案：

Q\A	1	2	3	4
1				√
2			√
3	√
4	√
5		√
6			√
7		√

More JOIN

1. List the films where the yr is 1962 [Show id, title]
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  SELECT id, title
  FROM movie
  WHERE yr=1962

Give year of ‘Citizen Kane’.

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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.
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  SELECT id, title, yr
  FROM movie
  WHERE title LIKE '%Star Trek%'
  ORDER BY yr

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

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SELECT id
 FROM actor
 WHERE name='Glenn Close'

What is the id of the film ‘Casablanca’

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SELECT id
 FROM movie
 WHERE title='Casablanca'

6.Obtain the cast list for ‘Casablanca’.

what is a cast list?
The cast list is the names of the actors who were in the movie.

Use movieid=11768, (or whatever value you got from the previous question)

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SELECT name
 FROM actor JOIN casting ON actorid=id
 WHERE movieid=11768

Obtain the cast list for the film ‘Alien’

SELECT DISTINCT name 
 FROM actor a JOIN casting c ON c.actorid=a.id
 WHERE c.movieid=(
  SELECT id FROM movie WHERE title='Alien'
 )

List the films in which ‘Harrison Ford’ has appeared

SELECT title
 FROM movie m JOIN casting c ON m.id=c.movieid
 WHERE c.actorid=(
  SELECT id FROM actor 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]
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  SELECT title
  FROM movie m JOIN casting c ON m.id=c.movieid
  WHERE c.actorid=(
  SELECT id FROM actor WHERE name='Harrison Ford'
  ) AND c.ord <> 1

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

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

Which were the busiest years for ‘John Travolta’, 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='John Travolta'
GROUP BY yr
HAVING COUNT(title)=(SELECT MAX(c) FROM
(SELECT yr,COUNT(title) AS c FROM
   movie JOIN casting ON movie.id=movieid
         JOIN actor   ON actorid=actor.id
 where name='John Travolta'
 GROUP BY yr) as t
)

1. 12.
  List the film title and the leading actor for all of the films ‘Julie Andrews’ played in.

Did you get “Little Miss Marker twice”?

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

Obtain a list, in alphabetical order, of actors who’ve had at least 30 starring roles.

SELECT name
FROM actor
  JOIN casting ON id = actorid 
  AND (SELECT COUNT(ord) FROM casting WHERE actorid = actor.id AND ord=1)>=30
  GROUP BY name

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

SELECT title, COUNT(actorid) as cast
FROM movie JOIN casting on id=movieid
WHERE yr = 1978
GROUP BY title
ORDER BY cast DESC, title

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

SELECT name
FROM actor JOIN casting ON actor.id=casting.actorid
WHERE movieid IN (
 SELECT movieid FROM casting JOIN actor ON casting.actorid=actor.id AND actor.name='Art Garfunkel'
) AND name <> 'Art Garfunkel'

Quiz MORE JOIN

选择题答案：

Q\A	2	3	4	5
1		√
2				√
3		√
4	√
5			√
6		√
7	√

USING NULL

List the teachers who have NULL for their department.

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SELECT name 
FROM teacher 
WHERE dept is NULL

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

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

Use a different JOIN so that all departments are listed.

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

1. 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’
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  SELECT teacher.name, COALESCE(mobile, '07986 444 2266')
  FROM teacher
1. 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.
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  SELECT teacher.name, COALESCE(dept.name, 'None')
  FROM teacher
  LEFT JOIN dept ON (teacher.dept = dept.id)
1. Use COUNT to show the number of teachers and the number of mobile phones.
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  SELECT COUNT(name), COUNT(mobile)
  FROM teacher
1. 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.
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  SELECT dept.name, COUNT(teacher.name)
  FROM teacher
  RIGHT JOIN dept ON (dept.id = teacher.dept)
  GROUP BY dept.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 teacher.name, 
CASE WHEN (teacher.dept = 1 or teacher.dept = 2) THEN 'Sci'
ELSE 'Art' END
FROM teacher

1. 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.
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  SELECT teacher.name,
  CASE
  WHEN (teacher.dept=1 OR teacher.dept=2) THEN 'Sci'
  WHEN teacher.dept=3 THEN 'Art'
  ELSE 'None' END
  FROM teacher

Using Null Quiz

选择题答案：

Q\A	1	2	3	4	5
1					√
2			√
3				√
4		√
5	√
6	√

Self join

这个有点难，从第5题开始我是参考别人的答案写的，我还不太理解，so bad！

1. How many stops are in the database.
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  SELECt COUNT(*) from stops

Find the id value for the stop ‘Craiglockhart’

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SELECT id
FROM stops
WHERE name='Craiglockhart'

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

1 2	SELECT id, name FROM stops JOIN route ON id=stop WHERE company='LRT' AND num=4

1. 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.
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  SELECT company, num, COUNT(*)
  FROM route WHERE stop=149 OR stop=53
  GROUP BY company, num
  HAVING COUNT(*) = 2
1. 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.
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  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)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
WHERE stopa.name='Haymarket' AND stopb.name='Leith'

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

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

Find the routes involving two buses that can go from Craiglockhart to Sighthill.
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.

SELECT DISTINCT bus1.num, bus1.company, name, bus2.num, bus2.company FROM (
SELECT start1.num, start1.company, stop1.stop FROM route AS start1 JOIN route AS stop1 
ON start1.num = stop1.num 
AND start1.company = stop1.company AND start1.stop != stop1.stop WHERE start1.stop = 
(SELECT id FROM stops WHERE name = 'Craiglockhart')) AS bus1
JOIN (SELECT start2.num, start2.company, start2.stop FROM route AS start2 
JOIN route AS stop2 ON start2.num = stop2.num 
AND start2.company = stop2.company 
AND start2.stop != stop2.stop 
WHERE stop2.stop = 
(SELECT id FROM stops WHERE name = 'Sighthill')) AS bus2 
ON bus1.stop = bus2.stop JOIN stops ON bus1.stop = stops.id

Self join Quiz

选择题答案：

Q\A	3	4	5
1	√
2			√
3		√