SQL Exercise: Country ID and number of cities in country has

SQL SORTING and FILTERING on HR Database: Exercise-25 with Solution

25. From the following table, write a SQL query to count the number of cities in each country. Return country ID and number of cities.

+-------------+------------------------------------------+-------------+---------------------+-------------------+------------+
| LOCATION_ID | STREET_ADDRESS                           | POSTAL_CODE | CITY                | STATE_PROVINCE    | COUNTRY_ID |
+-------------+------------------------------------------+-------------+---------------------+-------------------+------------+
|        1000 | 1297 Via Cola di Rie                     | 989         | Roma                |                   | IT         |
|        1100 | 93091 Calle della Testa                  | 10934       | Venice              |                   | IT         |
|        1200 | 2017 Shinjuku-ku                         | 1689        | Tokyo               | Tokyo Prefecture  | JP         |
|        1300 | 9450 Kamiya-cho                          | 6823        | Hiroshima           |                   | JP         |
|        1400 | 2014 Jabberwocky Rd                      | 26192       | Southlake           | Texas             | US         |
|        1500 | 2011 Interiors Blvd                      | 99236       | South San Francisco | California        | US         |
|        1600 | 2007 Zagora St                           | 50090       | South Brunswick     | New Jersey        | US         |
|        1700 | 2004 Charade Rd                          | 98199       | Seattle             | Washington        | US         |
|        1800 | 147 Spadina Ave                          | M5V 2L7     | Toronto             | Ontario           | CA         |
|        1900 | 6092 Boxwood St                          | YSW 9T2     | Whitehorse          | Yukon             | CA         |
|        2000 | 40-5-12 Laogianggen                      | 190518      | Beijing             |                   | CN         |
|        2100 | 1298 Vileparle (E)                       | 490231      | Bombay              | Maharashtra       | IN         |
|        2200 | 12-98 Victoria Street                    | 2901        | Sydney              | New South Wales   | AU         |
|        2300 | 198 Clementi North                       | 540198      | Singapore           |                   | SG         |
|        2400 | 8204 Arthur St                           |             | London              |                   | UK         |
|        2500 | Magdalen Centre, The Oxford Science Park | OX9 9ZB     | Oxford              | Oxford            | UK         |
|        2600 | 9702 Chester Road                        | 9629850293  | Stretford           | Manchester        | UK         |
|        2700 | Schwanthalerstr. 7031                    | 80925       | Munich              | Bavaria           | DE         |
|        2800 | Rua Frei Caneca 1360                     | 01307-002   | Sao Paulo           | Sao Paulo         | BR         |
|        2900 | 20 Rue des Corps-Saints                  | 1730        | Geneva              | Geneve            | CH         |
|        3000 | Murtenstrasse 921                        | 3095        | Bern                | BE                | CH         |
|        3100 | Pieter Breughelstraat 837                | 3029SK      | Utrecht             | Utrecht           | NL         |
|        3200 | Mariano Escobedo 9991                    | 11932       | Mexico City         | Distrito Federal, | MX         |
+-------------+------------------------------------------+-------------+---------------------+-------------------+------------+

Sample Solution:

-- Selecting 'country_id' and counting the number of records for each 'country_id' group
SELECT country_id, COUNT(*)
-- Specifying the table to retrieve data from ('locations')
FROM locations
-- Grouping the results by 'country_id'
GROUP BY country_id;

Sample Output:

 country_id | count
------------+-------
 CH         |     2
 MX         |     1
 US         |     4
 AU         |     1
 IT         |     2
 Ox         |     1
 JP         |     2
 CA         |     2
 DE         |     1
 NL         |     1
 SG         |     1
 CN         |     1
 UK         |     2
 IN         |     1
 BR         |     1
(15 rows)

Code Explanation:

The said query in SQL which aggregates data from the 'locations' table, grouping by the "country_id" column. It returns the values for each group of locations in the same country as below:
country_id: the country id of the locations
COUNT(*): the number of locations in the group/country.
Therefore the final result will be a list of country_id values and the count of locations for each country.

Relational Algebra Expression:

Relational Algebra Tree:

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