SQLZOO SELF JOIN

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: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: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'

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 JOIN route b

ON (a.num = b.num AND a.company = b.company)

WHERE a.stop = 115 and b.stop = 137

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 s1 ON a.stop = s1.id

JOIN stops s2 ON b.stop = s2.id

WHERE s1.name = 'Craiglockhart' 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 s1.name,a.company,a.num FROM route a JOIN route b

ON (a.company = b.company AND a.num = b.num)

JOIN stops s1 ON a.stop = s1.id

JOIN stops s2 ON b.stop = s2.id

WHERE s1.name = 'Craiglockhart' or s2.name = 'Craiglockhart'

10:Find the routes involving two buses that can go from Craiglockhart to Sighthill.

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 Sighthill, then join those on matching stops.

SELECT DISTINCT bus1.num, bus1.company, name, bus2.num, bus2.company FROM (

SELECT start1.num, start1.company, stop1.stop FROM route AS start1 JOIN route AS stop1

ON start1.num = stop1.num

AND start1.company = stop1.company AND start1.stop != stop1.stop WHERE start1.stop =

(SELECT id FROM stops WHERE name = 'Craiglockhart')) AS bus1 JOIN (SELECT start2.num, start2.company, start2.stop FROM route AS start2 JOIN route AS stop2 ON start2.num = stop2.num AND start2.company = stop2.company AND start2.stop != stop2.stop WHERE stop2.stop
= (SELECT id FROM stops WHERE name = 'Sighthill')) AS bus2 ON bus1.stop = bus2.stop JOIN stops ON bus1.stop = stops.id