SQLZoo答案

1 SELECT name

Some pattern matching queries

1. Find the country that start with Y

   SELECT name FROM world
    WHERE name LIKE 'F%'
   

2. Find the countries that end with y

   SELECT name FROM world
    WHERE name LIKE '%y'
   

3. Find the countries that contain the letter x

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

4. Find the countries that end with land

   SELECT name FROM world
     WHERE name LIKE '%land'
   

5. Find the countries that start with C and end with ia

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

6. Find the country that has oo in the name

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

7. Find the countries that have three or more a in the name

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

8. Find the countries that have “t” as the second character.

   SELECT name FROM world
    WHERE name LIKE '_t%'
   ORDER BY name
   

9. Find the countries that have two “o” characters separated by two others.

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

10. Find the countries that have exactly four characters.

    SELECT name FROM world
     WHERE name LIKE '____'
    

11. Find the country where the name is the capital city.

    SELECT name
      FROM world
     WHERE name = capital;
    

12. Find the country where the capital is the country plus “City”.

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

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

    select capital,name
    from world
    where capital like '%' + name + '%';
    

14. Find the capital and the name where the capital is an extension of name of the country.

    select capital,name
    from world
    where capital like name + '_%';
    

2 SELECT from WORLD Tutorial

use the SELECT command on the table world

1. show the name, continent and population of all countries.

   SELECT name, continent, population FROM world
   

2. 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
   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 'population / millions'
   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. 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 (more than 3 million) or big by population (more than 250 million) but not both. Show name, population and area.

   SELECT name, population, area
   FROM world
   WHERE 
    (area > 3000000 and population <= 250000000) or
    (area < 3000000 and population > 250000000)
   

9. For South America show population in millions and GDP in billions both to 2 decimal places.

   SELECT name, round(population/1000000, 2), round(gdp/1000000000, 2)
   FROM world
   WHERE continent = 'South America'
   

10. Show per-capita GDP for the trillion dollar countries to the nearest $1000.

    SELECT name, round(gdp/population, -3)
    FROM world
    WHERE gdp > 1000000000000
    

11. Show the name and capital where the name and the capital have the same number of characters.

    SELECT name, capital
     FROM world
     WHERE len(name) = len(capital)
    

12. 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.

    SELECT name, capital
    FROM world
    WHERE name <> capital and left(name, 1) = left(capital, 1)
    

13. Find the country that has all the vowels and no spaces in its name.

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

3 SELECT from Nobel Tutorial

We continue practicing simple SQL queries on a single table.

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 yr >= 2000 and subject = 'Peace'
   

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

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

9. 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')
   

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

11. Find all details of the prize won by PETER GRÜNBERG.(There contains an Non-ASCII characters)

    SELECT *
    FROM nobel
    WHERE winner = 'PETER GRÜNBERG'
    

12. Find all details of the prize won by EUGENE O’NEILL.(There contains an single qutoes)

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

13. 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
    

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. ...

4 SELECT within SELECT Tutorial

Looks at how we can use SELECT statements within SELECT statements to perform more complex queries.

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

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'))
   

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. 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),'%') 'percentage' 
   FROM world 
   WHERE continent='Europe' 
   

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

8. 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 y.continent = x.continent)
   

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 name, continent, population
   FROM world x
   WHERE 25000000 >= ALL(SELECT population
                      FROM world y
                      WHERE x.continent = y.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 x
    WHERE population >= ALL(SELECT 3*population 
                            FROM world y
                            WHERE x.continent = y.continent AND population > 0
                            AND y.name != x.name)
    

5. 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. The difference of the ‘HAVING’ and ‘WHERE’: The HAVING clause is tested after the GROUP BY. The WHERE filter takes place before the aggregating function.

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)
   FROM world
   WHERE continent = 'Africa'
   

4. How many countries have an area of at least 1000000

   SELECT COUNT(name)
   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) '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
   

5. The JOIN operation

This tutorial introduces JOIN which allows you to use data from two or more tables. The tables contain all matches and goals from UEFA EURO 2012 Football Championship in Poland and Ukraine.

1. 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. Show id, stadium, team1, team2 for just game 1012

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

3. 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 teamid='GER'
   

4. Show the team1, team2 and player for every goal scored by a player called Mario player LIKE ‘Mario%’

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

5. Show player, teamid, coach, gtime for all goals scored in the first 10 minutes gtime<=10

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

6. 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 JOIN eteam ON (team1=eteam.id)
   WHERE coach = 'Fernando Santos'
   

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

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

8. 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' OR team2='GER') AND teamid != 'GER')
   

9. Show teamname and the total number of goals scored.

   SELECT teamname, COUNT(player)
     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(player)
    FROM game JOIN goal ON(id=matchid)
    GROUP BY stadium
    

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

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

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

7. More JOIN operations

This tutorial introduces the notion of a join. The database consists of three tables movie , actor and casting .

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 name
   FROM movie JOIN casting ON (movie.id = movieid)
           JOIN actor ON (actor.id = actorid)
   WHERE title = 'Casablanca'
   

7. Obtain the cast list for the film ‘Alien’

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

8. List the films in which ‘Harrison Ford’ has appeared

   SELECT title
   FROM movie JOIN casting ON (movie.id = movieid)
           JOIN actor ON (actorid = actor.id)
   WHERE 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 title
   FROM movie JOIN casting ON (movie.id = movieid)
           JOIN actor ON (actorid = actor.id)
   WHERE name = 'Harrison Ford' AND ord != 1
   

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

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

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

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

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

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

    SELECT name
    FROM movie JOIN casting ON (movie.id = movieid)
           JOIN actor ON (actorid = actor.id)
    WHERE ord = 1
    GROUP BY name
    HAVING COUNT(name) >= 15 
    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(name)
    FROM movie JOIN casting ON (movie.id = movieid)
           JOIN actor ON (actorid = actor.id)
    WHERE yr = 1978
    GROUP BY title
    ORDER BY COUNT(name) DESC, title
    

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

    SELECT name
    FROM movie JOIN casting ON (movie.id = movieid)
           JOIN actor ON (actorid = actor.id)
    WHERE movieid in (SELECT movieid
                  FROM actor JOIN casting ON (actor.id = actorid)
                  WHERE name = 'Art Garfunkel') AND name != 'Art Garfunkel'
    

8. Using Null

We can use the phrase IS NULL to pick out fields. We can use IS NOT NULL similarly.

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. (This example show that what the effect of INNER JOIN looks like.)

   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. (This example show that what the effect of the left join looks like.)

   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. (This example show that what the effect of the right join looks like.)

   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.

   SELECT name, COALESCE(mobile, '07986 444 2266')
   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')
   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 teacher.dept = 1
         THEN 'Sci'
         WHEN teacher.dept = 2
         THEN 'Sci'
         ELSE'Art'
    END
   FROM teacher LEFT JOIN dept ON (teacher.dept = dept.id)
   

   Or we can use the combination of condition like the following answer:

   SELECT teacher.name,
    CASE WHEN teacher.dept = 1 OR teacher.dept = 2
         THEN 'Sci'
         ELSE'Art'
    END
   FROM teacher LEFT JOIN dept ON (teacher.dept = dept.id)
   

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 teacher.dept = 1 OR teacher.dept = 2
         THEN 'Sci'
         WHEN teacher.dept = 2
         THEN 'Art'
         ELSE'None'
    END
    FROM teacher LEFT JOIN dept ON (teacher.dept = dept.id)
    

8+. NSS Tutorial

1. Show the the percentage who STRONGLY AGREE

   SELECT A_STRONGLY_AGREE
     FROM nss
    WHERE question='Q01'
   AND institution='Edinburgh Napier University'
   AND subject='(8) Computer Science'
   

2. Show the institution and subject where the score is at least 100 for question 15.

   SELECT institution, subject
     FROM nss
    WHERE question='Q15'
   AND score >= 100
   

3. Show the institution and score where the score for ‘(8) Computer Science’ is less than 50 for question ‘Q15’

   SELECT institution, score
   FROM nss
   WHERE question='Q15'
     AND subject='(8) Computer Science'
     AND score < 50
   

4. Show the subject and total number of students who responded to question 22 for each of the subjects ‘(8) Computer Science’ and ‘(H) Creative Arts and Design’.

   SELECT subject, SUM(response)
     FROM nss
    WHERE question='Q22'
   AND (subject='(H) Creative Arts and Design'
        OR subject='(8) Computer Science')
   GROUP BY subject
   

5. Show the subject and total number of students who A_STRONGLY_AGREE to question 22 for each of the subjects ‘(8) Computer Science’ and ‘(H) Creative Arts and Design’.

   SELECT subject, SUM(response * A_STRONGLY_AGREE / 100)
     FROM nss
    WHERE question='Q22'
   AND (subject='(H) Creative Arts and Design'
       OR subject='(8) Computer Science')
   GROUP BY subject
   

9-. Window functions

9+. Window LAG

9. Self join

10.1 SELECT Quiz

1. Select the code which produces this table

   SELECT name, population
     FROM world
    WHERE population BETWEEN 1000000 AND 1250000
   

2. Pick the result you would obtain from this code:

   SELECT name, population
     FROM world
    WHERE name LIKE 'Al%'
   

   Answer: table-E

3. Select the code which shows the countries that end in A or L

   SELECT name FROM world
    WHERE name LIKE '%a' OR name LIKE '%l'
   

4. Pick the result from the query

   SELECT name,length(name)
   FROM world
   WHERE length(name)=5 and continent='Europe'
   

   Answer: table C

5. Here are the first few rows of the world table:. Pick the result you would obtain from this code: SELECT name, area*2 FROM world WHERE population = 64000 Answer: table c
6. Select the code that would show the countries with an area larger than 50000 and a population smaller than 10000000

   SELECT name, area, population
     FROM world
    WHERE area > 50000 AND population < 10000000
   

7. Select the code that shows the population density of China, Australia, Nigeria and France

   SELECT name, population/area
     FROM world
    WHERE name IN ('China', 'Nigeria', 'France', 'Australia')
   

10.2 BBC QUIZ

10.3 Nobel Quiz

10.4 Nested SELECT Quiz

10.5 SUM and COUNT Quiz

10.6 JOIN Quiz

10.7 JOIN Quiz 2

10.8 Using Null Quiz

10.9 Self join Quiz