Difference between revisions of "Self join"

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(Adding a hint to question 10)
 
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<p>[[Edinburgh_Buses. |Details of the database]] Looking at the data</p>
 
<p>[[Edinburgh_Buses. |Details of the database]] Looking at the data</p>
 
  stops('''id''', name)
 
  stops('''id''', name)
  route('''num''','''company''','''pros''', ''stop'')
+
  route('''num''','''company''','''pos''', ''stop'')
 +
 
 +
<div class='schema'></div>
 +
<div class = 'ref_section'>
 +
<table class = 'db_ref'>
 +
<tr><th>'''stops'''</th><th>'''route'''</th></tr>
 +
<tr><td>id</td><td>num</td></tr>
 +
<tr><td>name</td><td>company</td></tr>
 +
<tr><td></td><td>pos</td></tr>
 +
<tr><td></td><td>stop</td></tr>
 +
<tr><td></td><td></td></tr>
 +
</table>
 +
</div>
 +
 
 +
<div class="progress_panel"><div>
 +
  <div class="summary">Summary</div>
 +
  <div class="progressbarbg">
 +
    <div class="progressbar"></div>
 +
  </div>
 +
</div></div>
  
 
<div class='qu'>
 
<div class='qu'>
Line 60: Line 79:
  
 
<div class='qu'>
 
<div class='qu'>
Execute the self join shown and observe that b.stop gives all the places you can get to from Craiglockhart.
+
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.       
 
Change the query so that it shows the services from Craiglockhart to London Road.       
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
Line 107: Line 126:
  
 
<source lang='sql' class='ans'>
 
<source lang='sql' class='ans'>
SELECT R1.company, R1.num
+
SELECT DISTINCT R1.company, R1.num
 
   FROM route R1, route R2
 
   FROM route R1, route R2
 
   WHERE R1.num=R2.num AND R1.company=R2.company
 
   WHERE R1.num=R2.num AND R1.company=R2.company
Line 130: Line 149:
  
 
<div class='qu'>
 
<div class='qu'>
Give a list of the '''stops''' which may be reached from 'Craiglockhart' by taking one bus. Include the details of the appropriate service.         
+
Give a distinct list of the '''stops''' which may be reached from 'Craiglockhart' by taking one bus, including 'Craiglockhart' itself. Include the company and bus no. of the relevant services.         
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
</source>
 
</source>
  
 
<source lang='sql' class='ans'>
 
<source lang='sql' class='ans'>
SELECT S2.id, S2.name, R2.company, R2.num
+
SELECT DISTINCT S2.name, R2.company, R2.num
  FROM stops S1, stops S2, route R1, route R2
+
FROM stops S1, stops S2, route R1, route R2
  WHERE S1.name='Craiglockhart'
+
WHERE S1.name='Craiglockhart'
    AND S1.id=R1.stop
+
  AND S1.id=R1.stop
    AND R1.company=R2.company AND R1.num=R2.num
+
  AND R1.company=R2.company AND R1.num=R2.num
    AND R2.stop=S2.id
+
  AND R2.stop=S2.id
 
</source>
 
</source>
 
</div>
 
</div>
  
 
<div class='qu'>
 
<div class='qu'>
Show it is possinle to get from Sighthill to Craiglockhart.        
+
Find the routes involving two buses that can go from Craiglockhart to Sighthill.<br/>
 +
Show the bus no. and company for the first bus, the name of the stop for the transfer,<br/>
 +
and the bus no. and company for the second bus.
 +
<div class='hint' title='Hint'>Self-join twice to find buses that visit Craiglockhart and Sighthill, then join those on matching stops.</div>
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
</source>
 
</source>
 +
 +
<!-- the previous answer was much simpler, but it also didn't work? -->
  
 
<source lang='sql' class='ans'>
 
<source lang='sql' class='ans'>
select distinct a.name, c.name
+
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
from stops a JOIN route z ON a.id=z.stop
+
JOIN route y ON y.num = z.num
+
JOIN stops b ON y.stop=b.id
+
JOIN route x ON x.num = y.num
+
JOIN stops c ON c.id=x.stop
+
where a.name='Craiglockhart'
+
AND c.name ='Sighthill'
+
 
</source>
 
</source>
 
</div>
 
</div>
  
[[Self join Quiz]]
+
<div>
 +
<div class="lsclear">Clear your results</div>
 +
<p><div class="quizlink">[[Self join Quiz]]</div></p>
 +
</div>

Latest revision as of 00:44, 14 May 2014

[edit] Edinburgh Buses

Details of the database Looking at the data

stops(id, name)
route(num,company,pos, stop)
stopsroute
idnum
namecompany
pos
stop
Summary

How many stops are in the database.

 
SELECT COUNT(*) 
FROM stops

Find the id value for the stop 'Craiglockhart'

 
SELECT id 
FROM stops 
WHERE name='Craiglockhart'


Give the id and the name for the stops on the '4' 'LRT' service.

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

Routes and stops

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

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
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

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'
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

Give a list of all the services which connect stops 115 and 137 ('Haymarket' and 'Leith')

 
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

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

 
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'

Give a distinct list of the stops which may be reached from 'Craiglockhart' by taking one bus, including 'Craiglockhart' itself. Include the company and bus no. of the relevant services.

 
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
Clear your results

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