# Edinburgh Buses Table

| stops |
| ----- |
| *id*  |
| name  |

<br>

| route     |
| --------- |
| *num*     |
| *company* |
| *pos*     |
| stop      |

1. How many stops are in the database.&#x20;

`SELECT COUNT(*) FROM stops`

2. Find the id value for the stop 'Craiglockhart' (#53)

`SELECT id FROM stops WHERE name = 'Craiglockhart'`

3. Give the id and the name for the stops on the '4' 'LRT' service.&#x20;

```sql
SELECT id, name 
FROM stops s 
JOIN route r ON s.id = r.stop 
WHERE r.num = '4' and r.company = 'LRT'
```

4. 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.

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

5. 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. (#149)

```sql
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
```

6. 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'&#x20;

```sql
SELECT a.company, a.num, stops.name, stops.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'


```

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

```sql
SELECT DISTINCT a.company, a.num
FROM route a
JOIN route b 
ON a.company = b.company AND a.num = b.num
WHERE a.stop = 115 AND b.stop = 137
```

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

```sql
SELECT a.company, a.num
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 = 'Tollcross'

```

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

```sql
SELECT DISTINCT s.name, a.company, a.num
FROM route a 
JOIN route b ON (a.company=b.company AND a.num=b.num)
JOIN stops s ON (b.stop = s.id)
WHERE a.stop = 53 AND a.company = 'LRT'
```

10. &#x20;Find the routes involving two buses that can go from Craiglockhart to Lochend.\
    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.

```sql
SELECT 
    a.num, 
    a.company,  
    stops.name,  
    d.num, 
    d.company
FROM route a 
JOIN route b ON a.company = b.company AND a.num = b.num
JOIN stops ON b.stop = stops.id
JOIN route c ON c.stop = stops.id
JOIN route d ON c.company = d.company AND c.num = d.num
WHERE a.stop = (SELECT id 
                FROM stops
                WHERE name = 'Craiglockhart')
AND d.stop = (SELECT id 
             FROM stops 
             WHERE name = 'Lochend')
ORDER BY a.num, stops.name, d.num
```


---

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