Self join/zh

From SQLZoo

愛丁堡巴士

Details of the database Looking at the data

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

站(編號,名稱) 路線(號碼,巴士公司名,方向,站)

stopsroute
idnum
namecompany
pos
stop

數據庫中有多少個站stops

SELECT COUNT(*) 
FROM stops

找出車站 'Craiglockhart' 的 id

SELECT id 
FROM stops 
WHERE name='Craiglockhart'


列出巴士公司'LRT'的'4'號巴士線的站編號id 和 站名name

SELECT id, name FROM stops, route
  WHERE id=stop
    AND company='LRT'
    AND num='4'

Routes and stops

以下查詢列出途經 London Road (149) 或 Craiglockhart (53)的巴士線號碼。注意有兩條路線會經過這兩個站兩次。 加入 HAVING 語句來限制只列出這兩條路線。

SELECT company, num, COUNT(*)
FROM route WHERE stop=149 OR stop=53
GROUP BY company, num
SELECT company, num, COUNT(*)
FROM route WHERE stop=149 OR stop=53
GROUP BY company, num
HAVING COUNT(*)=2

執行自我合拼來,留意b.stop代表由Craiglockhart出發不用轉車可前住的地方。 修改它來顯示由Craiglockhart到 London Road的服務資料。

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
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

此題和上題相似,但是用兩個stops表來自我合拼。這樣我們可以用站名而非站編號。 修改它來顯示由Craiglockhart到 London Road的服務資料。 如你太悶,可試一試由 'Fairmilehead' 到 'Tollcross' (系統會當答錯的。正確有3條路線:11,15,315)

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'
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'

Using a self join

列出連接115 和 137 ('Haymarket' 和 'Leith') 的公司名和路線號碼。不要重覆。

SELECT DISTINCT R1.company, R1.num
  FROM route R1, route R2
  WHERE R1.num=R2.num AND R1.company=R2.company
    AND R1.stop=115 AND R2.stop=137

列出連接車站stops 'Craiglockhart' 到 'Tollcross' 的公司名和路線號碼。

SELECT R1.company, R1.num
  FROM route R1, route R2, stops S1, stops S2
  WHERE R1.num=R2.num AND R1.company=R2.company
    AND R1.stop=S1.id AND R2.stop=S2.id
    AND S1.name='Craiglockhart'
    AND S2.name='Tollcross'

不重覆列出可以由 'Craiglockhart' 乘一程車到達的站stops,包括'Craiglockhart'本身。 列出站名,公司名和路線號碼。

SELECT DISTINCT S2.name, R2.company, R2.num
FROM stops S1, stops S2, route R1, route R2
WHERE S1.name='Craiglockhart'
  AND S1.id=R1.stop
  AND R1.company=R2.company AND R1.num=R2.num
  AND R2.stop=S2.id

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.

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
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