SQLZOO

About

Welcome to the SQLzoo Solutions repository! This repository is dedicated to providing comprehensive and well-documented solutions to the exercises available on SQLzoo.net. The purpose of this repository is to serve as a learning resource for individuals looking to enhance their SQL skills, from beginners to more advanced users.

Purpose

The primary goals of this repository are:

Educational Resource: To offer clear, concise, and correct solutions to the various SQL challenges presented on SQLzoo.net, helping users to understand and learn SQL concepts effectively.

Reference Guide: To act as a handy reference for common SQL queries and techniques, showcasing practical examples and solutions that can be utilized in real-world scenarios.

Skill Development: To assist users in practicing and honing their SQL skills by providing step-by-step solutions and explanations for each exercise, covering a wide range of topics from basic SELECT statements to more complex operations like JOINs, window functions, and recursive queries.

Community Contribution: To encourage collaboration and knowledge sharing among the SQL community, allowing others to contribute their solutions, insights, and improvements. By exploring the solutions in this repository, you can deepen your understanding of SQL, learn new techniques, and prepare for data-related roles such as data analysts and database administrators.

Table of Content

SELECT BASICS
SELECT QUIZ
SELECT FROM WORLD
BBC QUIZ
SELECT FROM NOBEL TUTORIAL
NOBEL QUIZ
SELECT WITHIN SELECT TUTORIAL
NESTED SELECT QUIZ
SUM AND COUNT
SUM AND COUNT QUIZ
THE JOIN OPERATION
JOIN QUIZ
MORE JOIN OPERATIONS
MORE JOIN QUIZ
USING NULL
USING NULL QUIZ
Self Join
Self Join Quiz

SELECT BASICS

The example uses a WHERE clause to show the population of 'France'. Note that strings should be in 'single quotes'. Modify it to show the population of Germany:
```
SELECT population FROM world
WHERE name = 'Germany';
```
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');
```
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 QUIZ

Select the code which produces this table:

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

Pick the result you would obtain from this code:

SELECT name, population
FROM world
WHERE name LIKE "Al%";

Answer: Table-E

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';

Pick the result from the query:

-- Answer:
-- name	length(name)
-- Italy	5
-- Malta	5
-- Spain	5

Pick the result you would obtain from this code:
```
SELECT name, area*2 FROM world WHERE population = 64000;
```
Answer: Andorra 936
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;
```

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');

SELECT FROM WORLD

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;
```
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;
```
Give the name and the per capita GDP for those countries with a population of at least 200 million:
```
SELECT name, GDP/population as 'per capita GDP'
FROM world
WHERE population >= 200000000;
```
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 in m'
FROM world
WHERE continent = 'South America';
```

Show the name and population for France, Germany, Italy:

SELECT name, population
FROM world
WHERE name IN ('France', 'Germany', 'Italy');

Show the countries which have a name that includes the word 'United':
```
SELECT name
FROM world
WHERE name LIKE '%United%';
```
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;
```
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 XOR population > 250000000;
```
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';
```
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;
```
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);
```
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 name <> capital;
```
Equatorial Guinea and the 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:
```
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 '% %';
```

BBC QUIZ

Select the code which gives the name of countries beginning with U:
```
SELECT name
FROM world
WHERE name LIKE 'U%';
```
Select the code which shows just the population of the United Kingdom:
```
SELECT population
FROM world
WHERE name = 'United Kingdom';
```
Select the answer which shows the problem with this SQL code - the intended result should be the continent of France:
```
SELECT continent 
FROM world 
WHERE 'name' = 'France';
```
Answer: 'name' should be name
Select the result that would be obtained from the following code:
```
SELECT name, population / 10 
FROM world 
WHERE population < 10000;
```
Answer: Nauru/ 990
Select the code which would reveal the name and population of countries in Europe and Asia:
```
SELECT name, population
FROM world
WHERE continent IN ('Europe', 'Asia');
```

Select the code which would give two rows:

SELECT name FROM world
WHERE name IN ('Cuba', 'Togo');

Select the result that would be obtained from this code:

SELECT name FROM world
WHERE continent = 'South America'
AND population > 40000000;

Answer: Brazil/ Columbia

SELECT FROM NOBEL TUTORIAL

Change the query shown so that it displays Nobel prizes for 1950:
```
SELECT *
FROM nobel
WHERE yr = 1950;
```

Show who won the 1962 prize for Literature:

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

Show the year and subject that won 'Albert Einstein' his prize:

SELECT yr, subject
FROM nobel
WHERE winner = 'Albert Einstein';

Give the name of the 'Peace' winners since the year 2000, including 2000:

SELECT winner
FROM nobel
WHERE subject = 'Peace' AND yr >= 2000;

Show all details (yr, subject, winner) of the Literature prize winners for 1980 and 1984:
```
SELECT *
FROM nobel
WHERE subject = 'Literature' AND yr IN (1980, 1984);
```

Show all details of the presidential winners:

SELECT *
FROM nobel
WHERE winner LIKE 'President%';

Show the winners with first name John:

SELECT winner
FROM nobel
WHERE winner LIKE 'John%';

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);

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');

Show the year, subject, and name of winners for 1984 and 1980 for subjects not including Chemistry:

SELECT yr, subject, winner
FROM nobel
WHERE subject <> 'Chemistry' AND yr IN (1980, 1984);

NOBEL QUIZ

Select the code which gives the name of winners for 1950:
```
SELECT winner
FROM nobel
WHERE yr = 1950;
```

Select the code which shows just the subject for 1962 Literature prize winners:

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

Select the answer which shows the problem with this SQL code - the intended result should be the year and subject of Einstein's prize:
```
SELECT yr subject
FROM nobel
WHERE winner = 'Albert Einstein';
```
Answer: Missing comma between yr and subject.
Select the result that would be obtained from the following code:
```
SELECT winner
FROM nobel
WHERE subject = 'Peace' AND yr >= 2000;
```
Answer: The list of Peace winners since 2000.
Select the code which would reveal the year and subject for Literature prize winners for 1980 and 1984:
```
SELECT yr, subject
FROM nobel
WHERE subject = 'Literature' AND yr IN (1980, 1984);
```
Select the code which would give all details of the presidential winners:
```
SELECT *
FROM nobel
WHERE winner LIKE 'President%';
```
Select the result that would be obtained from this code:
```
SELECT yr, subject, winner
FROM nobel
WHERE winner LIKE 'John%';
```
Answer: The list of winners whose first name is John.

Select the code that shows 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);

Select the code which excludes Chemistry and Medicine winners for 1980:

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

Select the code that shows the year, subject, and name of winners for 1984 and 1980 excluding Chemistry:
```
SELECT yr, subject, winner
FROM nobel
WHERE subject <> 'Chemistry' AND yr IN (1980, 1984);
```

SELECT WITHIN SELECT TUTORIAL

List each country name where the population is larger than that of 'Russia'

SELECT name FROM world
WHERE population > (SELECT population FROM world
WHERE name = 'Russia');

Show the countries in Europe with a per capita GDP greater than 'United Kingdom'

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

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' OR name = 'Australia')
ORDER BY name;

Which country has a population that is more than United Kingdom but less than Germany? Show the name and the population.

SELECT name, population
FROM world
WHERE population > (SELECT population FROM world WHERE name = 'United Kingdom') 
    AND population < (SELECT population FROM world WHERE name = '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(CAST(ROUND(
    population/(SELECT population FROM world WHERE name = 'Germany')*100,0)AS CHAR), '%') AS percentage
    FROM world
    WHERE continent = 'Europe';

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');
```

Find the largest country (by area) in each continent, show the continent, the name, and the area:

SELECT continent, name, area
FROM world w1
WHERE area >= ALL (SELECT area FROM world WHERE continent = w1.continent);

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

SELECT continent, name
FROM world w1
WHERE name = (SELECT MIN(name) FROM world WHERE continent = w1.continent);

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 w1
WHERE 25000000 >= (SELECT MAX(population) FROM world WHERE continent = w1.continent);
```

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 w1
WHERE population > 3 * (SELECT MAX(population) 
    FROM world WHERE continent = w1.continent AND name <> w1.name);

NESTED SELECT QUIZ

Select the code that shows the name, region, and population of the smallest country in each region

SELECT region, name, population
FROM bbc x WHERE population <= ALL (SELECT population FROM bbc y WHERE y.region = x.region
    AND population > 0);

Select the code that shows the countries belonging to regions with all populations over 50000

SELECT name, region, population
FROM bbc x WHERE 50000 < ALL (SELECT population FROM bbc y WHERE x.region = y.region
    AND y.population > 0);

Select the code that shows the countries with a less than a third of the population of the countries around it

SELECT name, region FROM bbc x
WHERE population < ALL (SELECT population / 3 FROM bbc y WHERE y.region = x.region
    AND y.name != x.name);

Select the result that would be obtained from the following code:

SELECT name FROM bbc
WHERE population > (SELECT population FROM bbc WHERE name='United Kingdom')
AND region IN (SELECT region FROM bbc WHERE name = 'United Kingdom');

Answer: Table-D

Select the code that would show the countries with a greater GDP than any country in Africa (some countries may have NULL gdp values).

SELECT name FROM bbc
WHERE gdp > (SELECT MAX(gdp) FROM bbc WHERE region = 'Africa');

SELECT name FROM bbc
WHERE gdp > ALL (SELECT gdp FROM bbc WHERE region = 'Africa' AND gdp IS NOT NULL);

Select the code that shows the countries with population smaller than Russia but bigger than Denmark

SELECT name FROM bbc
WHERE population < (SELECT population FROM bbc WHERE name = 'Russia') AND
population > (SELECT population FROM bbc WHERE name = 'Denmark');

Select the result that would be obtained from the following code:

SELECT name FROM bbc
WHERE population > ALL (SELECT MAX(population) FROM bbc WHERE region = 'Europe')
AND region = 'South Asia';

Answer: Table-B

SUM and COUNT

Show the total population of the world.

SELECT SUM(population) AS 'Total Population'
FROM world;

List all the continents - just once each.
```
SELECT DISTINCT(continent)
FROM world;
```

Give the total GDP of Africa

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

How many countries have an area of at least 1000000

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

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

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

For each continent show the continent and number of countries.

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

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;
```

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

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

SUM and COUNT QUIZ

Select the statement that shows the sum of population of all countries in 'Europe'
```
SELECT SUM(population) FROM bbc WHERE region = 'Europe';
```
Select the statement that shows the number of countries with population smaller than 150000
```
SELECT COUNT(name) FROM bbc WHERE population < 150000;
```
Select the list of core SQL aggregate functions
```
AVG(), COUNT(), MAX(), MIN(), SUM()
```
Select the result that would be obtained from the following code:
```
SELECT region, SUM(area)
FROM bbc 
WHERE SUM(area) > 15000000 
GROUP BY region;
```
Answer: No result due to invalid use of the WHERE function
Select the statement that shows the average population of 'Poland', 'Germany' and 'Denmark'
```
SELECT AVG(population) FROM bbc WHERE name IN ('Poland', 'Germany', 'Denmark');
```

Select the statement that shows the medium population density of each region

SELECT region, SUM(population) / SUM(area) AS density FROM bbc GROUP BY region;

Select the statement that shows the name and population density of the country with the largest population

SELECT name, population / area AS density FROM bbc WHERE population = (SELECT MAX(population) FROM bbc);

Pick the result that would be obtained from the following code:

SELECT region, SUM(area) 
FROM bbc 
GROUP BY region 
HAVING SUM(area) <= 20000000;

Answer: Table-D

THE JOIN OPERATION

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';
```
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';
```
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 INNER JOIN goal ON (game.id = goal.matchid)
WHERE teamid = 'GER';
```
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 INNER JOIN goal ON (game.id = goal.matchid)
WHERE player LIKE 'Mario%';
```
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 INNER JOIN eteam ON (goal.teamid = eteam.id)
WHERE gtime <= 10;
```
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 INNER JOIN eteam ON (game.team1 = eteam.id)
WHERE coach = 'Fernando Santos';
```

List the player for every goal scored in a game where the stadium was 'National Stadium, Warsaw'

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

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 INNER JOIN goal ON (game.id = goal.matchid)
WHERE (team1 = 'GER' OR team2 = 'GER') AND teamid != 'GER';

Show teamname and the total number of goals scored.

SELECT stadium, COUNT(gtime)
FROM game INNER JOIN goal ON (game.id = goal.matchid)
GROUP BY stadium;

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

SELECT stadium, COUNT(gtime)
FROM game INNER JOIN goal ON (game.id = goal.matchid)
GROUP BY stadium;

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

SELECT id, mdate, COUNT(gtime)
FROM game INNER JOIN goal ON (game.id = goal.matchid)
WHERE (team1 = 'POL' OR team2 = 'POL')
GROUP BY id;

For every match where 'GER' scored, show matchid, match date and the number of goals scored by 'GER'

SELECT matchid, mdate, COUNT(gtime)
FROM game INNER JOIN goal ON (game.id = goal.matchid)
WHERE teamid = 'GER'
GROUP BY id;

SELECT matchid, mdate, COUNT(gtime)
FROM game INNER JOIN goal ON (game.id = goal.matchid)
GROUP BY id
HAVING teamid = 'GER';

This won't work as the column in the HAVING clause does not reference the column in the GROUP BY clause or the SELECT clause.

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.

JOIN QUIZ

You want to find the stadium where player 'Dimitris Salpingidis' scored. Select the JOIN condition to use:

Answer: game JOIN goal ON (id=matchid)
You JOIN the tables goal and eteam in an SQL statement. Indicate the list of column names that may be used in the SELECT line:

Answer: matchid, teamid, player, gtime, id, teamname, coach

Select the code which shows players, their team, and the number of goals they scored against Greece (GRE).

SELECT player, teamid, COUNT(*)
FROM game JOIN goal ON matchid = id
WHERE (team1 = 'GRE' OR team2 = 'GRE')
AND teamid != 'GRE'
GROUP BY player, teamid;

MORE JOIN OPERATIONS

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

Give year of 'Citizen Kane'.

SELECT yr
FROM movie
WHERE title = 'Citizen Kane';

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;
```

What id number does the actor 'Glenn Close' have?

SELECT id
FROM actor
WHERE name = 'Glenn Close';

What is the id of the film 'Casablanca'

SELECT id
FROM movie
WHERE title = 'Casablanca';

Obtain the cast list for 'Casablanca'. Use movieid=11768, (or whatever value you got from the previous question)
```
SELECT name
FROM actor INNER JOIN casting ON (actor.id = casting.actorid)
WHERE movieid = 11768;
```

Obtain the cast list for the film 'Alien'

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

List the films in which 'Harrison Ford' has appeared

SELECT movie.title
FROM casting INNER JOIN actor ON (casting.actorid = actor.id) INNER JOIN movie ON (casting.movieid = movie.id)
WHERE actor.name = 'Harrison Ford';

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 casting INNER JOIN actor ON (casting.actorid = actor.id) INNER JOIN movie ON (casting.movieid = movie.id)
WHERE (actor.name = 'Harrison Ford') AND (casting.ord <> 1);

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

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

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

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

SELECT title, 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 WHERE actorid IN (SELECT id FROM actor WHERE name = 'Julie Andrews')) AND ord = 1;

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

SELECT name
FROM actor INNER JOIN casting ON (actor.id = casting.actorid)
WHERE ord = 1
GROUP BY name
HAVING COUNT(movieid)>=15;

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

SELECT title, COUNT(actor.id)
FROM movie INNER JOIN casting ON (movie.id = casting.movieid)
        INNER JOIN actor ON (casting.actorid = actor.id)
WHERE yr = 1978
GROUP BY movie.id
ORDER BY COUNT(actor.id) DESC, title;

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

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

MORE JOIN QUIZ

Select the statement which lists the unfortunate directors of the movies which have caused financial loses (gross < budget)
```
SELECT name
FROM actor INNER JOIN movie ON actor.id = director
WHERE gross < budget
```

Select the correct example of JOINing three tables

SELECT *
FROM actor JOIN casting ON actor.id = actorid
JOIN movie ON movie.id = movieid

Select the statement that shows the list of actors called 'John' by order of number of movies in which they acted

SELECT name, COUNT(movieid)
FROM casting JOIN actor ON actorid=actor.id
WHERE name LIKE 'John %'
GROUP BY name ORDER BY 2 DESC

Select the result that would be obtained from the following code:

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

Answer: Table-B

Select the statement that lists all the actors that starred in movies directed by Ridley Scott who has id 351

SELECT name
FROM movie JOIN casting ON movie.id = movieid
JOIN actor ON actor.id = actorid
WHERE ord = 1 AND director = 351

There are two sensible ways to connect movie and actor. They are:

Answer: link the director column in movies with the primary key in actor, connect the primary keys of movie and actor via the casting table

Select the result that would be obtained from the following code:

SELECT title, yr 
FROM movie, casting, actor 
WHERE name='Robert De Niro' AND movieid=movie.id AND actorid=actor.id AND ord = 3

Answer: Table-B

USING NULL

List the teachers who have NULL for their department.
```
SELECT name
From teacher
WHERE dept IS NULL;
```
Note the INNER JOIN misses the teachers with no department and the departments with no teacher.
```
SELECT teacher.name, dept.name
FROM teacher JOIN dept
ON (teacher.dept=dept.id);
```

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);

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);

COALESCE: COALESCE takes any number of arguments and returns the first value that is not null.

    COALESCE(x,y,z) = x if x is not NULL
    COALESCE(x,y,z) = y if x is NULL and y is not NULL
    COALESCE(x,y,z) = z if x and y are NULL but z is not NULL
    COALESCE(x,y,z) = NULL if x and y and z are all NULL

SELECT name, party
      ,COALESCE(party,'None') AS aff
  FROM msp WHERE name LIKE 'C%';

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 mobile
FROM teacher;
```
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 name'
FROM teacher LEFT JOIN dept ON (teacher.dept = dept.id);
```

Use COUNT to show the number of teachers and the number of mobile phones.

SELECT COUNT(name) AS 'teacher count',
    COUNT(mobile) AS 'mobile count'
FROM teacher;

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;
```

CASE: CASE allows you to return different values under different conditions. If there no conditions match (and there is not ELSE) then NULL is returned.
```
CASE WHEN condition1 THEN value1
     WHEN condition2 THEN value2 
     ELSE def_value
END
```

SELECT name, population,
      CASE WHEN population<1000000
            THEN 'small'
            WHEN population<10000000
            THEN 'medium'
            ELSE 'large'
       END
FROM bbc

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.id = 1 THEN 'Sci'
         WHEN dept.id = 2 THEN 'Sci'
         ELSE 'Art'
    END
 FROM teacher LEFT JOIN dept ON (teacher.dept = dept.id);

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
FROM teacher LEFT JOIN dept ON (teacher.dept = dept.id);

USING NULL QUIZ

Select the code which uses an outer join correctly.

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

Select the correct statement that shows the name of department which employs Cutflower –

SELECT dept.name FROM teacher JOIN dept ON (dept.id = teacher.dept) WHERE teacher.name = 'Cutflower'

Select out of following the code which uses a JOIN to show a list of all the departments and number of employed teachers

SELECT dept.name, COUNT(teacher.name) FROM teacher RIGHT JOIN dept ON dept.id = teacher.dept GROUP BY dept.name

Using SELECT name, dept, COALESCE(dept, 0) AS result FROM teacher on teacher table will:

Answer: display 0 in result column for all teachers without department

Query:

SELECT name,
    CASE WHEN phone = 2752 THEN 'two'
         WHEN phone = 2753 THEN 'three'
         WHEN phone = 2754 THEN 'four'
         END AS digit
FROM teacher

Answer: 'four' for Throd

Select the result that would be obtained from the following code:

SELECT name, 
  CASE 
   WHEN dept 
    IN (1) 
    THEN 'Computing' 
   ELSE 'Other' 
  END 
FROM teacher

Answer: Table-A

Self Join

How many stops are in the database.

SELECT COUNT(id) as 'Stop Count'
 FROM stops;

Find the id value for the stop 'Craiglockhart'

SELECT id
FROM stops
WHERE name = 'Craiglockhart';

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';

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;
```
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;
```
The query shown is similar to the previous one, however by joining two copies of the stops table we can refer to stops by name rather than by number. Change the query so that the services between 'Craiglockhart' and 'London Road' are shown. If you are tired of these places try 'Fairmilehead' against 'Tollcross'
```
SELECT a.company, a.num, stopa.name, stopb.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
WHERE stopa.name='Craiglockhart' AND stopb.name = 'London Road';
```

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

SELECT a.company, a.num
 FROM route a JOIN route b ON a.num = b.num AND a.company = b.company
 WHERE a.stop = 115 AND b.stop = 137
 GROUP BY num, company;

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

SELECT a.company, b.num
FROM route a JOIN route b ON a.num = b.num AND a.company = b.company
JOIN stops startstop ON a.stop = startstop.id
JOIN stops endstop ON b.stop = endstop.id
WHERE startstop.name = 'Craiglockhart' AND endstop.name = 'Tollcross';

Give a distinct list of the stops which may be reached from 'Craiglockhart' by taking one bus, including 'Craiglockhart' itself, offered by the LRT company. Include the company and bus no. of the relevant services.

SELECT endstop.name, a.company, a.num
FROM route a JOIN route b ON a.num = b.num AND a.company = b.company
JOIN stops startstop ON a.stop = startstop.id
JOIN stops endstop ON b.stop = endstop.id
WHERE startstop.name = 'Craiglockhart' AND a.company = 'LRT';

Find the routes involving two buses that can go from Craiglockhart to Lochend. 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. Hint: Self-join twice to find buses that visit Craiglockhart and Lochend, then join those on matching stops.

SELECT X.num, X.company, X.name, Y.num, Y.company
FROM
    (SELECT start.num, start.company, transferstop.name
    FROM route start
    JOIN route mid ON start.num = mid.num AND start.company = mid.company
    JOIN stops startstop ON start.stop = startstop.id
    JOIN stops transferstop ON mid.stop = transferstop.id
    WHERE startstop.name = 'Craiglockhart') X
JOIN
    (SELECT end.num, end.company, transferstop.name
    FROM route mid
    JOIN route end ON mid.num = end.num AND mid.company = end.company
    JOIN stops transferstop ON mid.stop = transferstop.id
    JOIN stops endstop ON end.stop = endstop.id
    WHERE endstop.name = 'LOCHEND') Y
ON X.name = Y.name
ORDER BY X.num, X.name, Y.num;

-- MAKE SURE YOU DON'T JOIN NUM COMPANY FOR THE FIRST AND SECOND ROUTES AS THE TRANSFER FOR THE SECOND ROUTE WILL ONLY BE LIMITED TO THE NUMBER AND COMPANIES YOU USED ON YOUR FIRST TRIP!

Self Join Quiz

Select the code that would show it is possible to get from Craiglockhart to Haymarket

SELECT DISTINCT a.name, b.name
   FROM stops a JOIN route z ON a.id=z.stop
   JOIN route y ON y.num = z.num
   JOIN stops b ON y.stop=b.id
 WHERE a.name='Craiglockhart' AND b.name ='Haymarket';

Select the code that shows the stops that are on route.num '2A' which can be reached with one bus from Haymarket?

SELECT S2.id, S2.name, R2.company, R2.num
   FROM stops S1, stops S2, route R1, route R2
  WHERE S1.name='Haymarket' AND S1.id=R1.stop
    AND R1.company=R2.company AND R1.num=R2.num
    AND R2.stop=S2.id AND R2.num='2A'

Select the code that shows the services available from 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='Tollcross'