solutions.md

SQLZoo Solutions

Here are my solutions to SQL tutorials from SQLZoo

Sections

0. SELECT basics - Some simple queries to get you started
1. SELECT name - Some pattern matching queries
2. SELECT from World - In which we query the World country profile table.
3. SELECT from Nobel - Additional practice of the basic features using a table of Nobel Prize winners.
4. SELECT within SELECT - In which we form queries using other queries.
5. SUM and COUNT - In which we apply aggregate functions.
6. JOIN - In which we join two tables; game and goals.
7. More JOIN operations - In which we join actors to movies in the Movie Database.
8. using NULL - In which we look at teachers in departments. 8+. Numeric Examples - In which we look at a survey and deal with some more complex calculations. 9-. Window function - In which we examine UK general election results. 9+. COVID 19 - In which we measure the impact of COVID 19
9. Self join - In which we join Edinburgh bus routes to Edinburgh bus routes.

SELECT basics

1. 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'

2. 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' );

3. 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

SELECT name

1. 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%'

2. 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%'

4. 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%'
       AND NAME LIKE '%ia'

6. 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%'

7. 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%'

8. India and Angola have an n as the second character. You can use the underscore as a single character wildcard.

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

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

9. 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%'

10. Cuba and Togo have four characters names.

    Find the countries that have exactly four characters.

SELECT NAME
FROM   world
WHERE  Length(NAME) = 4

11. 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

12. 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')

13. 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, '%')

14. 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 Length(capital) > Length(NAME)

15. 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.

SELECT NAME,
       Replace(capital, NAME, '') ext
FROM   world
WHERE  capital LIKE Concat(NAME, '_%')

SELECT from World

1. 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. 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
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 (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

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.

SELECT NAME,
       Round(population / 1000000, 2),
       Round(gdp / 1000000000, 2)
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, -3)
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(NAME, 1) = LEFT(capital, 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 LIKE '%a%'
       AND NAME LIKE '%e%'
       AND NAME LIKE '%i%'
       AND NAME LIKE '%o%'
       AND NAME LIKE '%u%'
       AND NAME NOT LIKE '% %'

SELECT from Nobel

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 >= 1980
       AND yr <= 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  ( subject = 'Chemistry'
         AND yr = 1984 )
        OR ( subject = 'Physics'
             AND yr = 1980 )

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 GRÜNBERG

SELECT *
FROM   nobel
WHERE  winner = 'Peter Grünberg'

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,
          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.

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

SELECT within SELECT

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 = 'Argentina'), (SELECT continent
                                                    FROM   world
                                                    WHERE  NAME = '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), '%')
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  x.NAME <= ALL (SELECT NAME
                      FROM   world y
                      WHERE  x.continent = y.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
                               AND y.population > 0)

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

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

SUM and COUNT

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)
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

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'

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.

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 matchid from goal must match id 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.

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 team1,
       team2,
       player
FROM   game
       JOIN goal
         ON ( id = matchid )
WHERE  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 player,
       teamid,
       coach,
       gtime
FROM   goal
       JOIN eteam
         ON teamid = 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 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. 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'
          OR team2 = 'GER' )
       AND teamid != 'GER'

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

SELECT teamname,
       Count(teamname)
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(stadium)
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(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
       JOIN goal
         ON id = matchid
WHERE  ( 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.

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

More JOIN

12. Lead actor in Julie Andrews movies

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

13. Actors with 15 leading roles

SELECT NAME
FROM   actor
       JOIN casting
         ON ( id = actorid
              AND ord = '1' )
GROUP  BY NAME
HAVING ( Count(movieid) >= 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(actorid)
FROM   movie
       JOIN casting
         ON ( movie.id = movieid )
WHERE  yr = 1978
GROUP  BY title
ORDER  BY Count(actorid) DESC  

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

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

Using NULL

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')
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(teacher.NAME),
       Count(mobile)
FROM   teacher
       LEFT JOIN dept
              ON ( teacher.dept = dept.id )

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 IN ( 1, 2 ) THEN 'Sci'
           ELSE 'Art'
         END )
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 teacher.dept IN ( 1, 2 ) THEN 'Sci'
           WHEN teacher.dept = 3 THEN 'Art'
           ELSE 'None'
         END )
FROM   teacher

Numeric Examples

1. The example shows the number who responded for:

   • question 1
   • at 'Edinburgh Napier University'
   • studying '(8) Computer Science'

   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 score < 50
       AND subject = '(8) Computer Science'

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 IN ( '(8) Computer Science', '(H) Creative Arts and Design' )
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 IN ( '(8) Computer Science', '(H) Creative Arts and Design' )
GROUP  BY subject

6. Show the percentage of students who A_STRONGLY_AGREE to question 22 for the subject '(8) Computer Science' show the same figure for the subject '(H) Creative Arts and Design'.

   Use the ROUND function to show the percentage without decimal places.

SELECT subject,
       Round(Sum(response * a_strongly_agree) / Sum(response), 0)
FROM   nss
WHERE  question = 'Q22'
       AND subject IN ( '(8) Computer Science', '(H) Creative Arts and Design' )
GROUP  BY subject

7. Show the average scores for question 'Q22' for each institution that include 'Manchester' in the name.

   The column score is a percentage - you must use the method outlined above to multiply the percentage by the response and divide by the total response. Give your answer rounded to the nearest whole number.

SELECT institution,
       Round(Sum(response * score) / Sum(response), 0) score
FROM   nss
WHERE  question = 'Q22'
       AND ( institution LIKE '%Manchester%' )
GROUP  BY institution

8. Show the institution, the total sample size and the number of computing students for institutions in Manchester for 'Q01'.

SELECT institution,
       Sum(sample),
       Sum(CASE
             WHEN subject = '(8) Computer Science' THEN sample
           END) computing_students
FROM   nss
WHERE  question = 'Q01'
       AND ( institution LIKE '%Manchester%' )
GROUP  BY institution

Window function

1. Show the lastName, party and votes for the constituency 'S14000024' in 2017.

SELECT lastname,
       party,
       votes
FROM   ge
WHERE  constituency = 'S14000024'
       AND yr = 2017
ORDER  BY votes DESC

2. You can use the RANK function to see the order of the candidates. If you RANK using (ORDER BY votes DESC) then the candidate with the most votes has rank 1.

   Show the party and RANK for constituency S14000024 in 2017. List the output by party

SELECT party,
       votes,
       Rank()
         OVER (
           ORDER BY votes DESC) AS posn
FROM   ge
WHERE  constituency = 'S14000024'
       AND yr = 2017
ORDER  BY party

3. The 2015 election is a different PARTITION to the 2017 election. We only care about the order of votes for each year.

   Use PARTITION to show the ranking of each party in S14000021 in each year. Include yr, party, votes and ranking (the party with the most votes is 1).

SELECT yr,
       party,
       votes,
       Rank()
         OVER (
           partition BY yr
           ORDER BY votes DESC) AS posn
FROM   ge
WHERE  constituency = 'S14000021'
ORDER  BY party,
          yr

4. Edinburgh constituencies are numbered S14000021 to S14000026.

   *Use PARTITION BY constituency to show the ranking of each party in Edinburgh in 2017. Order your results so the winners are shown first, then ordered by constituency.

SELECT constituency,
       party,
       votes,
       Rank()
         OVER (
           partition BY constituency
           ORDER BY votes DESC) AS posn
FROM   ge
WHERE  constituency BETWEEN 'S14000021' AND 'S14000026'
       AND yr = 2017
ORDER  BY posn,
          constituency,
          votes DESC

5. You can use SELECT within SELECT to pick out only the winners in Edinburgh.

   Show the parties that won for each Edinburgh constituency in 2017.

SELECT constituency,
       party
FROM  (SELECT constituency,
              party,
              Rank()
                OVER (
                  partition BY constituency
                  ORDER BY votes DESC) AS posn
       FROM   ge
       WHERE  constituency BETWEEN 'S14000021' AND 'S14000026'
              AND yr = 2017) AS ed
WHERE  posn = 1

6. You can use COUNT and GROUP BY to see how each party did in Scotland. Scottish constituencies start with 'S'

   Show how many seats for each party in Scotland in 2017.

SELECT party,
       Count(1)
FROM   (SELECT constituency,
               party,
               Rank()
                 OVER (
                   partition BY constituency
                   ORDER BY votes DESC) AS posn
        FROM   ge
        WHERE  constituency LIKE 'S%'
               AND yr = 2017) AS ed
WHERE  posn = 1
GROUP  BY party

COVID 19

1. The example uses a WHERE clause to show the cases in 'Italy' in March 2020.

   Modify the query to show data from Spain

SELECT NAME,
       Day(whn),
       confirmed,
       deaths,
       recovered
FROM   covid
WHERE  NAME = 'Spain'
       AND Month(whn) = 3
       AND Year(whn) = 2020
ORDER  BY whn

Self join

1. How many stops are in the database.

SELECT Count (DISTINCT id)
FROM   stops

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
       JOIN route
         ON ( stops.id = route.stop )
WHERE  company = 'LRT'
       AND num = '4'

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(*)
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
       AND b.stop = 149

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 r1.company,
                r1.num
FROM   route r1,
       route r2
WHERE  r1.num = r2.num
       AND r1.company = r2.company
       AND r1.stop = 115
       AND r2.stop = 137

8. Give a list of the services which connect the stops 'Craiglockhart' and 'Tollcross'

SELECT DISTINCT r1.company,
                r1.num
FROM   route r1,
       route r2,
       stops s1,
       stops s2
WHERE  r1.num = r2.num
       AND r1.company = r2.company
       AND r1.stop = s1.id
       AND r2.stop = s2.id
       AND s1.NAME = 'Craiglockhart'
       AND s2.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 DISTINCT s2.NAME,
                r2.company,
                r2.num
FROM   route r1,
       route r2,
       stops s1,
       stops s2
WHERE  r1.num = r2.num
       AND r1.company = r2.company
       AND r1.stop = s1.id
       AND r2.stop = s2.id
       AND s1.NAME = 'Craiglockhart'