SQL Exercises: Display distinct salesman and their cities

Last update on March 27 2023 13:02:40 (UTC/GMT +8 hours)

SQL UNION: Exercise-2 with Solution

2. From the following tables, write a SQL query to find distinct salespeople and their cities. Return salesperson ID and city.

Sample table: Salesman

Sample table: Customer

Sample Solution:

SELECT salesman_id, city
FROM customer
UNION
(SELECT salesman_id, city
FROM salesman)

Sample Output:

salesman_id	city
5001	    New York
5002	    London
5002	    California
5006	    Paris
5007	    Rome
5002	    Paris
5005	    London
5003	    Berlin
5007	    Moscow
5003	    San Jose

Code Explanation:

The said query in SQL that retrieves a list of unique pairs of salesman_id and city from two different tables, customer and salesman. The UNION operator is used to combine the results of two separate SELECT statements into a single result set.
The resulting table will contain all the unique combinations of salesman_id and city that appear in either the customer or salesman tables.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Query Visualization:

Duration:

Rows:

Cost:

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Previous SQL Exercise: Display all salesmen and customer located in London.
Next SQL Exercise: Salesmen, customer involved in inventory management.

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SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

db=# SELECT * FROM xxx;
 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

PostgreSQL v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

Database: PostgreSQL

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

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