SQL Exercise: Company whose products have an average price
SQL JOINS: Exercise-24 with Solution
From the following tables write a SQL query to calculate and find the average price of items of each company higher than or equal to Rs. 350. 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
HAVING AVG(pro_price) >= 350;
Output of the Query:
avg | com_name -----------------------+---------- 5000.0000000000000000 | Samsung 650.0000000000000000 | iBall 1475.0000000000000000 | Epsion 500.0000000000000000 | Frontech 3200.0000000000000000 | Asus (5 rows)
Explanation:
The said SQL query is selecting the average price (AVG(pro_price)) of items and the name of the company (company_mast.com_name) they are associated with, by joining the item_mast and company_mast tables on the pro_com column in the item_mast table and the com_id column in the company_mast table. The results are grouped by the company name and only showing the results where the average price is greater than or equal to 350.
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Previous SQL Exercise: Display the average price of items of each company.
Next SQL Exercise: Display ID and price of most expensive product.
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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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