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

Last update on March 31 2023 06:07:48 (UTC/GMT +8 hours)

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.

Sample table : locations

Sample Solution:

SELECT country_id,  COUNT(*)  
	FROM locations 
		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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SQL: Tips of the Day

Difference between natural join and inner join

One significant difference between INNER JOIN and NATURAL JOIN is the number of columns returned-

Consider:

TableA                           TableB
+------------+----------+        +--------------------+    
|Column1     | Column2  |        |Column1  |  Column3 |
+-----------------------+        +--------------------+
| 1          |  2       |        | 1       |   3      |
+------------+----------+        +---------+----------+

The INNER JOIN of TableA and TableB on Column1 will return

SELECT * FROM TableA AS a INNER JOIN TableB AS b USING (Column1);
SELECT * FROM TableA AS a INNER JOIN TableB AS b ON a.Column1 = b.Column1;

+------------+-----------+---------------------+    
| a.Column1  | a.Column2 | b.Column1| b.Column3|
+------------------------+---------------------+
| 1          |  2        | 1        |   3      |
+------------+-----------+----------+----------+

The NATURAL JOIN of TableA and TableB on Column1 will return:

SELECT * FROM TableA NATURAL JOIN TableB
+------------+----------+----------+    
|Column1     | Column2  | Column3  |
+-----------------------+----------+
| 1          |  2       |   3      |
+------------+----------+----------+

Ref: https://bit.ly/3AG5CId

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