Self_JOIN.sql

-- 1)How many stops are in the database.
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 s.id, s.name 
FROM stops s 
	JOIN route r ON r.stop = s.id
WHERE r.num = 4 AND r.company = 'LRT'
ORDER BY r.pos;

-- 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) services between 'Craiglockhart' and 'London Road' are shown. If you are tired of these places try 'Fairmilehead' against 'Tollcross'
SELECT a.company, a.num, stopa.name, stopb.name
FROM route a 
	JOIN route b ON (a.company=b.company AND a.num=b.num)
	JOIN stops stopa ON (a.stop=stopa.id)
	JOIN stops stopb ON (b.stop=stopb.id)
WHERE stopa.name='Craiglockhart' AND stopb.name = 'London Road';

-- 7) Give a list of all the services which connect stops 115 and 137 ('Haymarket' and 'Leith')
SELECT DISTINCT a.company, a.num 
FROM route a
	RIGHT JOIN route b ON (a.company= b.company AND a.num = b.num)
WHERE a.stop = 115 AND b.stop = 137;
 
-- 8) Give a list of the services which connect the stops 'Craiglockhart' and 'Tollcross'
-- reccomanded 
SELECT DISTINCT a.company, a.num
FROM route a 
	RIGHT JOIN route b ON (a.company = b.company AND a.num = b.num)
	JOIN stops ac ON ac.id = a.stop
	JOIN stops bc ON bc.id = b.stop
WHERE ac.name = 'Craiglockhart' AND bc.name = 'Tollcross';
-- or
SELECT DISTINCT a.company, a.num 
FROM route a 
	RIGHT JOIN route b ON (a.company = b.company AND a.num = b.num)
WHERE a.stop = (SELECT id FROM stops WHERE name = 'Craiglockhart') AND b.stop = (SELECT id FROM stops WHERE name = 'Tollcross');

-- 9)Give a distinct list of the stops which may be reached from 'Craiglockhart' by taking one bus, including 'Craiglockhart' itself, offered by the LRT company. Include the company and bus no. of the relevant services.
SELECT DISTINCT bc.name, b.company, b.num 
FROM route a 
	JOIN route b ON (a.company = b.company AND a.num = b.num)
	JOIN stops ac ON ac.id = a.stop
	JOIN stops bc ON bc.id = b.stop
WHERE ac.name = 'Craiglockhart';

-- 10) 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.
SELECT 
    x.num, x.company, x.name, y.num, y.company
FROM
    (SELECT DISTINCT
        bc.name, b.company, b.num
    FROM
        route a
    JOIN route b ON (a.company = b.company AND a.num = b.num)
    JOIN stops ac ON ac.id = a.stop
    JOIN stops bc ON bc.id = b.stop
    WHERE
        ac.name = 'Craiglockhart') x
        JOIN
    (SELECT DISTINCT
        cc.name, c.company, c.num
    FROM
        route c
    JOIN route d ON (c.company = d.company AND c.num = d.num)
    JOIN stops cc ON cc.id = c.stop
    JOIN stops dc ON dc.id = d.stop
    WHERE
        dc.name = 'Lochend') y ON (y.name = x.name)
ORDER BY x.num , name , y.num