09. Self JOIN.sql

/*
1. How many stops are in the database.
*/

SELECT count(*)
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 stops.id
	,stops.NAME
FROM stops
INNER JOIN route ON stops.id = route.stop
WHERE route.company = 'LRT'
	AND route.num = '4'

/*
Routes and stops
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
INNER JOIN route b ON (
		a.company = b.company
		AND a.num = b.num
		)
WHERE a.stop = 53
	AND b.stop = (
		SELECT stops.id
		FROM stops
		WHERE stops.NAME = 'London Road'
		)

/*
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
INNER JOIN route b ON (
		a.company = b.company
		AND a.num = b.num
		)
INNER JOIN stops stopa ON (a.stop = stopa.id)
INNER JOIN stops stopb ON (b.stop = stopb.id)
WHERE stopa.NAME = 'Craiglockhart'
	AND stopb.NAME = 'London Road'

/*
Using a self join
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
INNER JOIN route b ON (
		a.company = b.company
		AND a.num = b.num
		)
INNER JOIN stops stopa ON (a.stop = stopa.id)
INNER JOIN stops stopb ON (b.stop = stopb.id)
WHERE stopa.NAME = 'Haymarket'
	AND stopb.NAME = 'Leith'

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

SELECT DISTINCT a.company
	,a.num
FROM route a
INNER JOIN route b ON (
		a.company = b.company
		AND a.num = b.num
		)
INNER JOIN stops stopa ON (a.stop = stopa.id)
INNER JOIN stops stopb ON (b.stop = stopb.id)
WHERE stopa.NAME = 'Craiglockhart'
	AND stopb.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 stopb.NAME
	,b.company
	,b.num
FROM route a
INNER JOIN route b ON (
		a.num = b.num
		AND a.company = b.company
		)
INNER JOIN stops stopa ON (a.stop = stopa.id)
INNER JOIN stops stopb ON (b.stop = stopb.id)
WHERE stopa.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 rx.num
	,rx.company
	,sx.NAME AS change_at
	,ry.num
	,ry.company
FROM route rx
INNER JOIN route ry ON (rx.stop = ry.stop)
INNER JOIN stops sx ON (sx.id = rx.stop)
WHERE rx.num != ry.num
	AND rx.company IN (
		SELECT DISTINCT ra.company
		FROM route ra
		INNER JOIN stops sa ON (
				ra.stop = (
					SELECT id
					FROM stops sa
					WHERE NAME = 'Craiglockhart'
					)
				)
		WHERE ra.num = rx.num
		)
	AND ry.company IN (
		SELECT DISTINCT rb.company
		FROM route rb
		INNER JOIN stops sb ON (
				rb.stop = (
					SELECT id
					FROM stops sb
					WHERE NAME = 'Lochend'
					)
				)
		WHERE rb.num = ry.num
		);