SQL Exercise: Display the average price of items of each company

Last update on February 07 2023 09:47:19 (UTC/GMT +8 hours)

SQL JOINS: Exercise-23 with Solution

From the following tables write a SQL query to calculate the average price of items of each company. Return average value and company name.

Sample table: company_mast

Sample table: item_mast

Sample Solution:

SELECT AVG(pro_price), company_mast.com_name
   FROM item_mast INNER 
     JOIN company_mast
        ON item_mast.pro_com= company_mast.com_id
           GROUP BY company_mast.com_name;

Output of the Query:

avg			com_name
5000.0000000000000000	Samsung
650.0000000000000000	iBall
1475.0000000000000000	Epsion
500.0000000000000000	Frontech
250.0000000000000000	Zebronics
3200.0000000000000000	Asus

Explanation:

The said SQL query is selecting the average price of all products (AVG(pro_price)) and the company name (com_name) from the tables item_mast and company_mast, and grouping the results by the company name.
The query is using an INNER JOIN to combine the data from the two tables, linking the 'pro_com' column in item_mast to the 'com_id' column in company_mast.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Query Visualization:

Duration:

Rows:

Cost:

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Previous SQL Exercise: Display the item name, price, and company name.
Next SQL Exercise: Company whose products have an average price.

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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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