SQL Exercise: Display each item producer company
SQL JOINS: Exercise-21 with Solution
From the following tables write a SQL query to select all rows from both participating tables as long as there is a match between pro_com and com_id.
Sample table: company_mast
Sample table: item_mast
Sample Solution:
SELECT *
FROM item_mast
INNER JOIN company_mast
ON item_mast.pro_com= company_mast.com_id;
Output of the Query:
pro_id pro_name pro_price pro_com com_id com_name 101 Mother Board 3200.00 15 15 Asus 102 Key Board 450.00 16 16 Frontech 103 ZIP drive 250.00 14 14 Zebronics 104 Speaker 550.00 16 16 Frontech 105 Monitor 5000.00 11 11 Samsung 106 DVD drive 900.00 12 12 iBall 107 CD drive 800.00 12 12 iBall 108 Printer 2600.00 13 13 Epsion 109 Refill cartridge 350.00 13 13 Epsion 110 Mouse 250.00 12 12 iBall
Explanation:
The said SQL query that retrieves all columns from the table item_mast and company_mast and combines the data based on the join condition specified in the ON clause.
The join condition is that the 'pro_com' column in the item_mast table matches the 'com_id' column in the company_mast table. The result of the query is a new table that contains all the columns from both item_mast and company_mast, with rows that have matching values in the specified columns.
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Previous SQL Exercise: Customers from a different city than the salesmen.
Next SQL Exercise: Display the item name, price, and company name.
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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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