SQL Exercises: Customer and salesmen who lives in the same city

Last update on March 27 2023 12:43:03 (UTC/GMT +8 hours)

SQL Query on Multiple Tables: Exercise-1 with Solution

From the following tables, write a SQL query to find the salespeople and customers who live in the same city. Return customer name, salesperson name and salesperson city.

Sample table: salesman

Sample table: customer

Sample Solution:

SELECT customer.cust_name,
salesman.name, salesman.city
FROM salesman, customer
WHERE salesman.city = customer.city;

Output of the query:

cust_name	name		city
Nick Rimando	James Hoog	New York
Brad Davis	James Hoog	New York
Julian Green	Pit Alex	London
Fabian Johnson	Mc Lyon		Paris
Fabian Johnson	Nail Knite	Paris
Brad Guzan	Pit Alex	London

Code Explanation:

The said query in SQL that joins the 'salesman' and 'customer' tables based on the city column. The result set includes the customer name (cust_name), salesman name (name), and city from the salesman table. The WHERE clause specifies the join condition between the two tables, which is that the city column must be equal in both tables.

Relational Algebra Expression:

Relational Algebra Tree:

Explanation:

Syntax to find the customer and salesmen who lives in same city

Visual presentation:

Result to the customer and salesmen who lives in same city

Practice Online

Query Visualization:

Duration:

Rows:

Cost:

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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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