Working EnvironmentData and Question From SQLZoo (http://sqlzoo.net/wiki/SELECT_from_WORLD_Tutorial)
Data : http://sqlzoo.net/
MySQL 5.7.21
GUI : HeidiSQL 9.5.0.5196
OS : Windows 10 v1702 64 bit
text editor : Atom Version 1.23.3
WORLD
name continent area population gdp
Afghanistan Asia 652230 25500100 20343000000
Albania Europe 28748 2831741 12960000000
Algeria Africa 2381741 37100000 188681000000
Andorra Europe 468 78115 3712000000
Angola Africa 1246700 20609294 100990000000
...
NOBEL
yr subject winner
1960 Chemistry Willard F. Libby
1960 Literature Saint-John Perse
1960 Medicine Sir Frank Macfarlane Burnet
1960 Medicine Peter Madawar
...
Game
id mdate stadium team1 team2
1001 8 June 2012 National Stadium, Warsaw POL GRE
1002 8 June 2012 Stadion Miejski (Wroclaw) RUS CZE
1003 12 June 2012 Stadion Miejski (Wroclaw) GRE CZE
1004 12 June 2012 National Stadium, Warsaw POL RUS
...
Goal
matchid teamid player gtime
1001 POL Robert Lewandowski 17
1001 GRE Dimitris Salpingidis 51
1002 RUS Alan Dzagoev 15
1002 RUS Roman Pavlyuchenko 82
...
Eteam
id teamname coach
POL Poland Franciszek Smuda
RUS Russia Dick Advocaat
CZE Czech Republic Michal Bilek
GRE Greece Fernando Santos
...
Movie
id title yr director budget gross
10003 "Crocodile" Dundee II 1988 38 15800000 239606210
10004 'Til There Was You 1997 49 10000000
...
Actor
id name
20 Paul Hogan
50 Jeanne Tripplehorn
...
Casting
movieid actorid ord
10003 20 4
10004 50 1
Teacher
id dept name phone mobile
101 1 Shrivell 2753 07986 555 1234
102 1 Throd 2754 07122 555 1920
103 1 Splint 2293
104 Spiregrain 3287
105 2 Cutflower 3212 07996 555 6574
106 Deadyawn 3345
...
dept
id name
1 Computing
2 Design
3 Engineering
...
Edinburgh Buses.
stops
Field Type Notes
id INTEGER Arbitrary value
name CHAR(30) The name of an area served by at least one bus
route
Field Type Notes
num CHAR(5) The number of the bus - as it appears on the front of the vehicle. Oddly these numbers often include letters
company CHAR(3) Several bus companies operate in Edinburgh. The main one is Lothian Region Transport - LRT
pos INTEGER This indicates the order of the stop within the route. Some routes may revisit a stop. Most buses go in both directions.
stop INTEGER This references the stops table
QA - 1
1.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
2.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
3. Give the name and the per capita GDP for those countries with a population of at least 200 million.
select name,gdp/population as pc_gdp
from world
where population > 200000000
4. 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 'Population (M)'
from world
where continent = 'South America'
5. Show the name and population for France, Germany, Italy
select name, population
from world
where name in ('France','Germany','Italy')
6. Show the countries which have a name that includes the word ‘United’
select name
from world
where name like 'United%'
7. 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
8. Exclusive OR (XOR). Show the countries that are big by area or big by population 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 and population > 250000000) or (area > 3000000 and population < 250000000)
9. 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
select name,round(population/1000000,2) as 'Population(M)',round(gdp/1000000000,2) as 'GDP(B)'
from world
where continent = 'South America'
10. 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/1000,0) * 1000
from world
where gdp > 1000000000000
11. 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)
12. 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(capital,1) = left(name,1) and (capital != name)
13. 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 not like '%a%e%i%o%u%' and name not like '% %'
QA - 2
1. Change the query shown so that it displays Nobel prizes for 1950.
SELECT yr, subject, winner
FROM nobel
WHERE yr = 1950
2. Show who won the 1962 prize for Literature.
SELECT winner
FROM nobel
WHERE yr = 1962
AND subject = 'Literature'
3. Show the year and subject that won ‘Albert Einstein’ his prize.
select yr,subject
from nobel
where winner = 'Albert Einstein'
4. Give the name of the ‘Peace’ winners since the year 2000, including 2000.
select winner
from nobel
where subject = 'Peace' and yr >= 2000
5. 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
6. 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')
7. Show the winners with first name John
select winner
from nobel
where winner like 'John%'
8. Show the year, subject, and name of Physics winners for 1980 together with the Chemistry winners for 1984.
select *
from nobel
where (yr = 1984 and subject = 'Chemistry') or (yr = 1980 and subject = 'Physics')
9. Show the year, subject, and name of winners for 1980 excluding Chemistry and Medicine
select *
from nobel
where yr = 1980 and subject not in ('Chemistry','Medicine')
10. 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 *
from nobel
where (subject = 'Medicine' and yr < 1910) or
(subject = 'Literature' and yr >=2004)
11. Find all details of the prize won by PETER GR¨¹NBERG
This word can’t show up on my Atom utf-8, but works fine on html
select *
from nobel
where winner = 'PETER GR¨¹NBERG'
12. Find all details of the prize won by EUGENE O’NEILL
select *
from nobel
where winner = 'EUGENE O''NEILL'
13. Knights in order 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
14. 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 ('Chemistry','Physics'), subject,winner
QA - 3
1.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')
2.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')
3.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
4.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')
5.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((population/(select population from world where name = 'Germany')* 100),0),'%') as Pop_perc
from world
where continent = 'Europe'
order by Pop_perc
6.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 > all(
select gdp from world where continent = 'Europe' and gdp > 0)
7.Find the largest country (by area) in each continent, show the continent, the name and the area:
select continent,area
from world as w1
where area = (select max(area)
from world as w2
where w1.continent = w2.continent)
8.List each continent and the name of the country that comes first alphabetically.
select continent,name
from world as w1
where name = (
select name from world as w2 where w1.continent = w2.continent
order by name limit 1)
9.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 as w1
where 25000000 > all(
select w2.population from world as w2 where w1.continent = w2.continent )
10.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 as w1
where population > all(
select 3*population from world as w2 where w1.continent = w2.continent and w1.name != w2.name)
QA - 4
1. How many stops are in the database.
select count(distinct stop)
from route
2. Find the id value for the stop ‘Craiglockhart’
select id
from stops
where name = 'Craiglockhart'
3. Give the id and the name for the stops on the ‘4’ ‘LRT’ service.
select id,name
from stops inner join route
on stops.id = route.stop
where num = 4 and company = 'LRT'
4. 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( * ) as N
FROM route
WHERE stop=149 OR stop=53
GROUP BY company, num
having count(* ) = 2
5. 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
6. 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'
7. Give a list of all the services which connect stops 115 and 137 (‘Haymarket’ and ‘Leith’)
select distinct a.company,a.num
from route as a inner join route as b
on a.company = b.company and a.num = b.num
where a.stop = 115 and b.stop = 137
8. Give a list of the services which connect the stops ‘Craiglockhart’ and ‘Tollcross’
select distinct a.company,a.num
from route as a inner join route as b
on a.company = b.company and a.num = b.num
where a.stop = (select id from stops where name = 'Craiglockhart')
and b.stop = (select id from stops where name = 'Tollcross')
9. 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 stops.name, a.company,a.num
from route as a inner join route as b
on a.company = b.company and a.num = b.num
inner join stops
on stops.id = b.stop
where a.stop = (select id from stops where name = 'Craiglockhart')
and a.company = 'LRT'
10. 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 ta.num,ta.company,ta.name,tb.num,tb.company
from
(select a.num,a.company,b.stop,stops.name
from route as a inner join route as b
on a.company = b.company and a.num = b.num
inner join stops on stops.id = b.stop
where a.stop = (select id from stops where name = 'Craiglockhart')) as ta
inner join
(select a.num,a.company,a.stop
from route as a inner join route as b
on a.company = b.company and a.num = b.num
where b.stop = (select id from stops where name = 'Sighthill')) as tb
on ta.stop = tb.stop
QA - 5
1. Show the total population of the world.
SELECT SUM(population)
FROM world
2. List all the continents - just once each.
select distinct continent
from world
3. Give the total GDP of Africa
select sum(gdp) as 'Total GDP'
from world
where continent = 'Africa'
4. How many countries have an area of at least 1000000
select count(name) as 'Number of Countries'
from world
where area > 1000000
5. What is the total population of (‘Estonia’, ‘Latvia’, ‘Lithuania’)
select sum(population)
from world
where name in ('Estonia', 'Latvia', 'Lithuania')
6. For each continent show the continent and number of countries.
select continent,count(name) as 'Number of Countries'
from world
group by continent
7. 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
8. List the continents that have a total population of at least 100 million.
select continent
from world
group by continent
having sum(population) > 100000000
QA - 6
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’
select matchid,player
from goal
where teamid = 'GER'
2. 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
3. You can combine the two steps into a single query with a JOIN. 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)
SELECT player,teamid,stadium,mdate
FROM game JOIN goal ON (id=matchid)
where teamid = 'Ger'
4. 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 ga.team1,ga.team2,go.player
from game as ga inner join goal as go
on ga.id = go.matchid
where go.player like 'Mario%'
5. 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 go.player, go.teamid, e.coach, go.gtime
FROM goal as go inner join eteam as e
on go.teamid = e.id
WHERE gtime<=10
6. 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.
select ga.mdate,e.teamname
from game as ga inner join eteam as e
on ga.team1 = e.id
where e.coach = 'Fernando Santos'
7. List the player for every goal scored in a game where the stadium was ‘National Stadium, Warsaw’
select go.player
from goal as go inner join game as ga
on go.matchid = ga.id
where ga.stadium = 'National Stadium, Warsaw'
8. 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 ((team1='GER' and team2!='GER') or (team1!='GER' and team2='GER')) and goal.teamid != 'GER'
9. Show teamname and the total number of goals scored.
SELECT teamname, count(teamid)
FROM eteam JOIN goal ON id=teamid
group by teamname
10. Show the stadium and the number of goals scored in each stadium.
select stadium,count(teamid)
from game as ga inner join goal as go
on ga.id = go.matchid
group by stadium
11. For every match involving ‘POL’, show the matchid, date and the number of goals scored.
SELECT matchid,mdate, count(goal.teamid)
FROM game JOIN goal ON matchid = id
WHERE (team1 = 'POL' OR team2 = 'POL')
group by matchid,mdate
12. For every match where ‘GER’ scored, show matchid, match date and the number of goals scored by ‘GER’
select matchid,mdate,count(teamid)
from game as ga inner join goal as go
on ga.id = go.matchid
where (team1 = 'GER' or team2 = 'GER') and teamid = 'GER'
group by matchid,mdate
13. 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 JOIN goal ON matchid = id
group by mdate,team1,team2
order by mdate,matchid,team1,team2
QA - 7
1. List the teachers who have NULL for their department.
select name
from teacher
where dept is null
2. 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)
3. Use a different JOIN so that all teachers are listed.
select teacher.name,dept.name
from teacher left join dept
on teacher.dept = dept.id
4. Use a different JOIN so that all departments are listed.
select teacher.name,dept.name
from teacher right join dept
on teacher.dept = dept.id
5. 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 Number
from teacher
6. 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 teacher.name,coalesce(dept.name,'None') as DEPT
from teacher left join dept
on teacher.dept = dept.id
7. Use COUNT to show the number of teachers and the number of mobile phones.
select count(name),count(mobile)
from teacher
8. 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 dept.name,count(teacher.name)
from teacher right join dept
on teacher.dept = dept.id
group by dept.name
9. 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 dept in (1,2) then 'Sci' else 'Art' end) as D
from teacher
10. 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 teacher.name,
(case when dept in (1,2) then 'Sci' when dept = 3 then 'Art' else 'None' end) as D
from teacher
QA - 8
1. List the films where the yr is 1962 [Show id, title]
SELECT id, title
FROM movie
WHERE yr=1962
2. Give year of ‘Citizen Kane’.
select yr
from movie
where title = 'Citizen Kane'
3. 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
4. What id number does the actor ‘Glenn Close’ have?
select id
from actor
where name = 'Glenn Close'
5. What is the id of the film ‘Casablanca’
select id
from movie
where title = 'Casablanca'
6. Obtain the cast list for ‘Casablanca’.
select distinct name
from actor inner join casting
on casting.actorid = actor.id
where movieid = 11768
7. Obtain the cast list for the film ‘Alien’
select actor.name
from (actor inner join casting
on actor.id = casting.actorid) inner join movie
on movie.id = casting.movieid
where movie.title = 'Alien'
8. List the films in which ‘Harrison Ford’ has appeared
select movie.title
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where actor.name = 'Harrison Ford'
9. 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 movie.title
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where actor.name = 'Harrison Ford' and casting.ord != 1
10. List the films together with the leading star for all 1962 films.
select movie.title,actor.name
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where movie.yr = '1962' and casting.ord = 1
11. 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
)
12. List the film title and the leading actor for all of the films ‘Julie Andrews’ played in.
select movie.title,actor.name
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where movieid in (select casting.movieid
from casting inner join actor
on casting.actorid = actor.id
where actor.name = 'Julie Andrews') and casting.ord = 1
13. Obtain a list, in alphabetical order, of actors who’ve had at least 30 starring roles.
select name
from(
select count(movie.id) as N,name
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where casting.ord = 1
group by name
order by N desc) as temp
where temp.N >= 30
order by name
14. 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 N
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where yr = 1978
group by title
order by N desc,title
15. List all the people who have worked with ‘Art Garfunkel’.
select distinct actor.name
from (movie inner join casting
on movie.id = casting.movieid) inner join actor
on casting.actorid = actor.id
where movieid in(select movieid
from casting inner join actor
on casting.actorid = actor.id
where name = 'Art Garfunkel')
and name != 'Art Garfunkel'
