9 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 id, name
FROM stops JOIN route ON id = stop
WHERE num = '4' AND company = 'LRT'


--4. 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(*) > 1


--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. 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 a.company, a.num
FROM route a JOIN route b ON (a.company=b.company AND a.num=b.num)
WHERE a.stop=115 AND b.stop = 137
GROUP BY a.company, a.num


--8. Give a list of the services which connect the stops 'Craiglockhart' and 'Tollcross'
SELECT a.company, a.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='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 stopb.name, a.company, a.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 = 'Craiglockhart' 
GROUP BY stopb.name, a.company, a.num, stopa.name


--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
--Not my solution
SELECT a.num, a.company, stopsa.name, d.num, d.company
FROM route a JOIN route b ON (a.num = b.num AND a.company = b.company)
JOIN stops stopsa ON b.stop = stopsa.id
JOIN route c ON stopsa.id = c.stop
JOIN route d ON (c.num = d.num) AND (c.company = d.company)
WHERE (a.stop = (SELECT id FROM stops WHERE name = 'Craiglockhart')) AND
      (d.stop = (SELECT id FROM stops WHERE name = 'Lochend'))
ORDER BY a.num, stopsa.name, d.num;