SQL ZOO练习——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 = route.stop)
WHERE company = 'LRT' AND num = 4

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 num = 4 OR num = 45

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.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'

SELECT a1.company, a1.num
FROM route a1
JOIN route a2 ON (a1.company = a2.company AND a1.num = a2.num)
JOIN stops stopa ON (a1.stop = stopa.id)
JOIN stops stopb ON (a2.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 stopa.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 stopb.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.

SELECT DISTINCT bus1.num, bus1.company, name, bus2.num, bus2.company
FROM (SELECT a.num, a.company, b.stop
      FROM route a
      JOIN route b ON a.num = b.num AND a.company = b.company
        AND a.stop != b.stop
      WHERE a.stop = (SELECT id FROM stops
                      WHERE name =  'Craiglockhart')) bus1
JOIN (SELECT a2.num, a2.company, a2.stop
      FROM route a2
      JOIN route b2 ON a2.num = b2.num AND a2.company = b2.company
      AND a2.stop != b2.stop
      WHERE b2.stop = (SELECT id 
                       FROM stops
                       WHERE name = 'Sighthill')) bus2
ON bus1.stop = bus2.stop
JOIN stops ON bus1.stop = stops.id

-- 这题很复杂
-- 参考链接：https://zhuanlan.zhihu.com/p/38488508