SQLZoo-answers

My answers for SQLZoo tutorials questions

Sections:

1. SELECT basics
2. SELECT from WORLD
3. SELECT from NOBEL (harder questions)
4. NESTED SELECT
5. SUM and COUNT
6. JOIN
7. More JOIN
8. Using NULL
9. Self JOIN

Additional material:

1. Using GROUP BY and HAVING

SELECT basics

Some simple queries to get you started

1.1 Modify it to show the population of Germany

  SELECT population FROM world
    WHERE name = 'Germany'

1.2 Modify it to show the population of Russia

  SELECT population FROM world
    WHERE name = 'Russia'

2.1 Show the name and the population for 'Sweden', 'Norway' and 'Denmark'.

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

2.2 Show the name and the population for 'Brazil', 'Russia', 'India' and 'China'.

SELECT name, population FROM world
  WHERE name IN ('Brazil', 'Russia', 'India', 'China');

3.1 Shows countries with an area of 250,000-300,000 sq. km.

SELECT name, area FROM world
  WHERE area BETWEEN 250000 AND 300000

3.2 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 from WORLD

1.1 Show the name, continent and population of all countries.

SELECT name, continent, population FROM world

1.2 Show the name and population with limit 5 countries.

SELECT name, population FROM world
  LIMIT 5

2.1 Show the name for the countries that have a population of at least 200,000,000 (200 million).

SELECT name FROM world
  WHERE population >= 200000000

2.2 Show the name and population for the countries that have a population of at less 1,000,000 (1 million).

SELECT name, population FROM world
  WHERE population <= 1000

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')

6.1 Show the countries which have a name that includes the word 'United'

SELECT name
FROM world
WHERE name LIKE 'United%'

6.2 Show name, continent for all countries which have a name continent equals 'Asia'

SELECT name, continent
FROM world
WHERE continent = 'Asia'

6.3 Show unique continent from table world, and order by continent

SELECT DISTINCT continent
FROM world
ORDER BY continent

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. 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. Exclusive OR (XOR).

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.

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.

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

11.1 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 )

11.2 Show the name, continent and capital and number of characters, where name countries like 'G%'

SELECT name,      LENGTH(name),
       continent, LENGTH(continent),
       capital, LENGTH(capital)
  FROM world
 WHERE name LIKE 'G%'

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. You can use the function LEFT to isolate the first character. You can use <> (or !=) as the NOT EQUALS operator.

SELECT name, capital
FROM world
WHERE LEFT(name,1) = LEFT(capital,1) 
  AND name != capital

13. Find the country that has all the vowels (a e i o u) and no spaces in its name. You can use the phrase name NOT LIKE '%a%' to exclude characters from your results.

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

SELECT from NOBEL (harder questions)

1.1 Shown so that it displays Nobel prizes for 1950.

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

1.2 Show all winners Nobel laureates from 1950-1952 years, and order it's by years. SELECT yr, subject, winner FROM nobel WHERE yr BETWEEN 1950 AND 1952 ORDER BY yr 1.3 Show unique subject from Nobel tables

SELECT DISTINCT subject
FROM nobel

2.1 Show who won the 1962 prize for Literature.

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

2.2 Show who won the 1960 prize for Physics.

SELECT winner
FROM nobel
WHERE yr = 1960
AND subject = 'Physics'

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 * (yr, subject, winner -> another variant)
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 yr, subject, winner
FROM nobel
WHERE winner IN ('Theodore Roosevelt','Woodrow Wilson', 'Jimmy Carter', 'Barack Obama')

7.1 Show the winners with first name John

SELECT winner
FROM nobel
WHERE winner LIKE 'John%'

7.2 Show the winners with name consist 'stian'

SELECT winner
FROM nobel
WHERE winner LIKE '%stian%'

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

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

SELECT *
FROM nobel
WHERE subject NOT IN ('Chemistry', 'Medicine')
AND yr = 1980

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 (Non-ASCII characters)

SELECT *
FROM nobel
WHERE winner LIKE '%Grünberg%'

12. Find all details of the prize won by EUGENE O'NEILL

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. Направление сортировки: ASC (по возрастанию) или DESC (по убыванию)

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

NESTED SELECT

1.1 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')

1.2 Another variant

SELECT name, population
FROM world
WHERE name = 'Russia' //population = 146745098

SELECT name
FROM world
WHERE population > 146745098

2. Show the countries in Europe with a per capita GDP greater than 'United Kingdom'. Per Capita GDP.

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 ('South America', 'Oceania')
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.
   The format should be Name, Percentage for example:

---

|name | percentage |

| Albania | 3% |
| Andorra | 0% |
|....

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

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

SELECT continent, name
FROM world as x
WHERE name = (SELECT name FROM world as y WHERE x.continent = y.continent 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 name, continent, population
FROM world as x
WHERE continent != ALL(SELECT continent FROM world WHERE population > 25000000)

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

- NONE

SUM and COUNT

0.1. The total population and GDP of Europe.

SELECT SUM(population), SUM(gdp)
  FROM bbc
  WHERE region = 'Europe'

0.2. What are the regions?

SELECT DISTINCT region FROM bbc

0.3. Show the name and population for each country with a population of more than 100000000. Show countries in descending order of population.

SELECT name, population
  FROM bbc
  WHERE population > 100000000
  ORDER BY population DESC

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

1.1 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

SELECT * FROM goal 
  WHERE player LIKE '%Bender'

1.2 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. 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 goal JOIN game ON (goal.matchid=game.id)
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. 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. More difficult questions
   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. (COUNT and GROUP BY)
   You should COUNT(*) in the SELECT line and GROUP BY teamname

SELECT teamname, COUNT(teamid/gtime) as goals
FROM eteam JOIN goal ON id=teamid
GROUP BY teamname

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

SELECT stadium, COUNT(matchid) as goals
FROM game JOIN goal ON id=matchid
WHERE game.id=goal.matchid
GROUP BY stadium

10.2 Another answer

SELECT stadium,COUNT(1)
  FROM goal JOIN game 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(id)
FROM goal JOIN game 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 goal JOIN game ON matchid=id
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. 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, matchid, team1, team2

More JOIN

More details about the database.

1. 1962 movies. 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. Star Trek movies. 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. id for actor Glenn Close. What id number does the actor 'Glenn Close' have?

SELECT id
FROM actor
WHERE NAME = 'Glenn Close'

5. id for Casablanca
   What is the id of the film 'Casablanca'

SELECT id
FROM movie
WHERE title = 'Casablanca'

6. Cast list for Casablanca 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)

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

7. Alien cast list.
   Obtain the cast list for the film 'Alien'

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

8.1 Harrison Ford movies. List the films in which 'Harrison Ford' has appeared

SELECT title
FROM movie, casting, actor
WHERE name='Harrison Ford'
   AND movieid=movie.id
   AND actorid=actor.id

8.2 Another variant

SELECT title
FROM movie
WHERE id IN (SELECT movieid FROM casting WHERE actorid = 2216)

9. Harrison Ford as a supporting actor.
   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, casting, actor
WHERE movieid = movie.id
   AND actorid = actor.id
   AND name = 'Harrison Ford'
   AND ord != 1

10. Lead actors in 1962 movies.
    List the films together with the leading star for all 1962 films.

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

11.1 Busy years for Rock Hudson.
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

11.2 Another variant

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

12. Lead actor in Julie Andrews movies.
    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 JOIN casting ON (movieid = movie.id 
                            AND ord = 1)
           JOIN actor ON (actorid = actor.id)
WHERE movie.id IN (
           SELECT movieid FROM casting
             WHERE actorid IN (
               SELECT id FROM actor
                 WHERE name = 'Julie Andrews'))

14.1 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, casting, actor
WHERE yr = 1978
   AND movieid = movie.id
   AND actorid = actor.id
GROUP BY title
ORDER BY COUNT(actorid) DESC, title

14.2 Another variant

SELECT title, COUNT(actorid)
FROM casting,movie                
WHERE yr=1978
      AND movieid=movie.id
GROUP BY title
ORDER BY 2 DESC,1 ASC

Using NULL

1. List the teachers who have NULL for their department

SELECT name
FROM teacher
WHERE dept IS NULL

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

Self JOIN

More details about the database.

1.1 How many stops are in the database.

SELECT COUNT(*) 
FROM stops

1.2 Another variant

SELECT COUNT(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, route
WHERE id = stop
 AND company = 'LRT'
 AND num = '4'

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

SELECT company, num, COUNT(*) aaa
FROM route WHERE stop=149 OR stop=53
GROUP BY company, num
HAVING aaa = 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'
//WHERE stopa.name='Fairmilehead' AND stopb.name='Tollcross'

More about Self-Join

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

SELECT DISTINCT b.company , b.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 = 'Leith' AND stopb.name = 'Haymarket'

7.2 Another variant

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

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

SELECT R1.company, R1.num
FROM route R1, route R2
WHERE R1.num=R2.num AND R1.company=R2.company
AND R1.stop=53 AND R2.stop=230

8.2 Another variant

SELECT 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'

8.3 Another variant

SELECT 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=(SELECT name FROM stops WHERE id = 53) 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'

group-by-and-having

1. For each continent show the number of countries:

SELECT continent, COUNT(name)
FROM world
GROUP BY continent

2. For each continent show the total population:

SELECT continent, SUM(population)
FROM world
GROUP BY continent

3. WHERE and GROUP BY. The WHERE filter takes place before the aggregating function.
   For each relevant continent show the number of countries that has a population of at least 200000000.

SELECT continent, COUNT(name)
FROM world
WHERE population > 200000000
GROUP BY continent

4. GROUP BY and HAVING. The HAVING clause is tested after the GROUP BY.
   You can test the aggregated values with a HAVING clause.
   Show the total population of those continents with a total population of at least half a billion.

SELECT continent, SUM(population)
FROM world
GROUP BY continent
HAVING SUM(population) > 500000000