Difference between revisions of "Self join"

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(As stated, Question 7 result should only give one line for each unique pair of company, num; thus added DISTINCT.)
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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="summary">Summary</div>
+
<div class='schema'></div>
−
 
+
<div class = 'ref_section'>
−
<div class="progressbarbg">
+
<table class = 'db_ref'>
−
  <div class="progressbar"></div>
+
<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>
 +
 +
<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'>
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<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'>
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<source lang='sql' class='ans'>
 
<source lang='sql' class='ans'>
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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
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<div class='qu'>
 
<div class='qu'>
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Show it is possible to get from Sighthill to Craiglockhart.         
+
Try to show it is possible to get from Sighthill to Craiglockhart by changing lanes twice.
 +
This DB doesn't make this easy!          
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
</source>
 
</source>
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</div>
 
</div>
  
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[[Self join Quiz]]
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<div>
 +
<div class="lsclear">Clear your results</div>
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<p><div class="quizlink">[[Self join Quiz]]</div></p>
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</div>

Revision as of 20:40, 31 March 2013

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 list of the stops which may be reached from 'Craiglockhart' by taking one bus. Include the details of the appropriate service. 

 
SELECT S2.id, 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

Try to show it is possible to get from Sighthill to Craiglockhart by changing lanes twice. This DB doesn't make this easy! 

 
SELECT DISTINCT a.name, c.name
 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'
Clear your results

Self join Quiz

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