SQL Exercise: Display ID and price of most expensive product
SQL JOINS: Exercise-25 with Solution
From the following tables write a SQL query to find the most expensive product of each company. Return pro_name, pro_price and com_name.
Sample table: company_mast
Sample table: item_mast
Sample Solution:
SELECT A.pro_name, A.pro_price, F.com_name
FROM item_mast A INNER JOIN company_mast F
ON A.pro_com = F.com_id
AND A.pro_price =
(
SELECT MAX(A.pro_price)
FROM item_mast A
WHERE A.pro_com = F.com_id
);
Output of the Query:
pro_name pro_price com_name Monitor 5000.00 Samsung DVD drive 900.00 iBall Printer 2600.00 Epsion ZIP drive 250.00 Zebronics Mother Board 3200.00 Asus Speaker 550.00 Frontech
Explanation:
The said SQL query is selecting the product name (A.pro_name), product price (A.pro_price) and the company name (F.com_name) of an item, by joining the item_mast table A and company_mast table F on the pro_com column of the item_mast table and the com_id column of the company_mast table. It is also using a subquery to find the highest price of item of the same company, and then join it with the outer query to show the product name, product price, and the company name that has the highest price.
Practice Online
Query Visualization:
Duration:

Rows:

Cost:

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Previous SQL Exercise: Company whose products have an average price.
Next SQL Exercise: Display employees including their department.
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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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