sqlzoo习题答案--Self join

Details of the database Looking at the data

stops(id, name)
route(num,company,pos, stop)

stops	route
id	num
name	company
	pos
	stop

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 id,name
from stops join route
on stops.id=route.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')

*****方法一*****

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

*****方法二*****

select distinct a.company,a.num
from route a inner join route b 
on a.num=b.num and a.company=b.company and a.stop=115 and b.stop=137

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

select distinct a.company,a.num
from route a join route b
on (a.company=b.company and a.num=b.num)
join stops stopa on stopa.id=a.stop
join stops stopb on stopb.id=b.stop
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 s1.name,r1.company,r1.num 
from stops s1,stops s2,route r1,route r2
where s2.id=r2.stop
 and  s2.name='Craiglockhart'
 and  r2.company=r1.company
 and  r2.num=r1.num
 and  r1.stop=s1.id

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.

*****草稿*****

select r1.num,r1.stop from route r1
where r1.num in(select distinct r1.num
from route r1 join stops stopa
on r1.stop=stopa.id
where name='Craiglockhart')



select r2.num,r2.stop from route r2
where r2.num in(
select distinct r2.num
from route r2 join stops stopb
on r2.stop=stopb.id
where name='Sighthill')





select r1.num,r1.company,stopc.name,r2.num,r2.company from route r1,route r2,stops stopc
where r1.num in(select distinct r1.num
from route r1 join stops stopa
on r1.stop=stopa.id
where name='Craiglockhart')
and r2.num in(
select distinct r2.num
from route r2 join stops stopb
on r2.stop=stopb.id
where name='Sighthill')
and r1.num!=r2.num
and r1.stop=stopc.id
and r1.stop=stopa.id

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