Difference between revisions of "Self join"
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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''',''' | + | route('''num''','''company''','''pos''', ''stop'') |
| − | <div class= | + | <div class='schema'></div> |
| − | + | <div class = 'ref_section'> | |
| − | <div class= | + | <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> | ||
| + | |||
| + | <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 66: | 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 151: | Line 164: | ||
<div class='qu'> | <div class='qu'> | ||
| − | Show it is possible to get from Sighthill to Craiglockhart. | + | Show it is possible to get from Sighthill to Craiglockhart by changing lanes twice. |
<source lang='sql' class='def'> | <source lang='sql' class='def'> | ||
</source> | </source> | ||
| Line 167: | Line 180: | ||
</div> | </div> | ||
| − | [[Self join Quiz]] | + | <div> |
| + | <div class="lsclear">Clear your results</div> | ||
| + | <p><div class="quizlink">[[Self join Quiz]]</div></p> | ||
| + | </div> | ||
Revision as of 11:31, 13 February 2013
Edinburgh Buses
Details of the database Looking at the data
stops(id, name) route(num,company,pos, stop)
| stops | route |
|---|---|
| id | num |
| name | company |
| pos | |
| stop | |
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 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
Show it is possible to get from Sighthill to Craiglockhart by changing lanes twice.
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'
