找到涉及兩輛可能從A到B的巴士的路線

Question

從sqlzoo到SQL的問題： 查找涉及兩輛可從Craiglockhart前往Sighthill的巴士的路線。 顯示巴士號碼。公司爲第一輛公交車，名稱爲轉車站， 和公交車號。第二輛公共汽車的公司。

這是我找到的代碼，但它不會工作：

SELECT DISTINCT a.num, a.company, 
      trans1.name , c.num, c.company 
FROM route a JOIN route b 
ON (a.company = b.company AND a.num = b.num) 
JOIN (route c JOIN route d ON (c.company = d.company AND c.num= d.num)) 
JOIN stops start ON (a.stop = start.id) 
JOIN stops trans1 ON (b.stop = trans1.id) 
JOIN stops trans2 ON (c.stop = trans2.id) 
JOIN stops end ON (d.stop = end.id) 
WHERE start.name = 'Craiglockhart' AND end.name = 'Sighthill' 
      AND trans1.name = trans2.name 
ORDER BY a.num ASC , trans1.name 

Answer 1

我假設一個路由由一個簡單的開始（這是pos屬性）和結束，這是stop屬性。希望這可以讓你指向正確的方向。外部查詢獲取在Craiglockhart中啓動的所有路由，並且內部查詢獲取以Sighthill結尾的所有路由，然後您加入第一個路由的結束與第二個路由的開始相同的兩個路由。

SELECT r1.Num  AS FirstBusNo, 
      r1.Company AS FirstBusCompany, 
      Stops.Name AS Transfer, 
      r2.Num  AS SecondBusNo, 
      r2.Company AS SecondBusCompany 
     FROM Route r1 
INNER JOIN Stops trans 
     ON r1.Stop = trans.Id 
INNER JOIN (SELECT Num, Company, Pos 
      FROM Route, Stops 
      WHERE Id = stop 
      AND Name = 'Sighthill') r2 
     ON r1.stop = r2.pos 
    WHERE Stops.Name = 'Craiglockhart'; 

Answer 2

當你遇到這樣的大問題時，最好的做法是對其進行分區。從技術上講，所有出發Craiglockhart的公共汽車最終應達到Sighthill，並有足夠的轉移，但我們只限於1次轉移（因爲這是問題的原因）。

所以基本上，你需要找到一輛從Craiglockhart出發的公共汽車，以及一輛到達Sighthill的公共汽車，並且你需要找到它們之間的所有相交站點。

第2個部分是非常容易的（和你有那部分右）：

SELECT DISTINCT num , company 
FROM route 
INNER JOIN stops ON stop = id 
WHERE name='Craiglockhart' 

和

SELECT DISTINCT num, company AS bus 
FROM route 
INNER JOIN stops ON stop = id 
WHERE name='Sighthill' 

從那裏，你有離開Craiglockhart和所有所有公交車的列表巴士到達Sighthill。你唯一存在的問題是找到那兩個相交的地方。

該解決方案的初始部分同樣簡單;您需要獲取2輛巴士的名稱以及停靠站的名稱。

SELECT DISTINCT r1.num as NoFrom, r1.company CoFrom, name,r2.num as NoTo, r2.company CoTo 

這意味着你必須查詢至少3個表

FROM stops 
INNER JOIN route r1 ON r1.stop = id 
INNER JOIN route r2 ON r2.stop = id 

所以，現在，你收集了所有可能的轉會名單，你只需要過濾掉無用的。一個次優的方式做這將是這樣的：

WHERE exists(
    SELECT 1 
    FROM route r3 
    INNER JOIN stops s1 ON r3.stop = s1.id 
    WHERE s1.name='Craiglockhart' AND r3.num = r1.num AND r3.company = r1.company) 

AND exists(
    SELECT 1 
    FROM route r4 
    INNER JOIN stops s2 ON r4.stop = s2.id 
    WHERE s1.name='Sighthill' AND r4.num = r2.num AND r4.company = r2.company) 

所以基本上做的一種方式將是：

SELECT DISTINCT r1.num as NoFrom, r1.company CoFrom, name, r2.num as NoTo, r2.company CoTo 
FROM stops 
INNER JOIN route r1 ON r1.stop = id 
INNER JOIN route r2 ON r2.stop = id 
WHERE exists(
    SELECT 1 
    FROM route r3 
    INNER JOIN stops s1 ON r3.stop = s1.id 
    WHERE s1.name='Craiglockhart' AND r3.num = r1.num AND r3.company = r1.company) 
AND exists(
    SELECT 1 
    FROM route r4 
    INNER JOIN stops s2 ON r4.stop = s2.id 
    WHERE s2.name='Sighthill' AND r4.num = r2.num AND r4.company = r2.company) 

Relevant SQLFiddle

有一件事值得一提的是，雖然我用「當存在（...）」時，你的練習明確暗示你應該使用自我連接。如果你想學習，我不會給你全部的答案，所以一定要去嘗試把它轉換成自己的聯接;）