SQL Exercise: A list of which salesmen work for which customers
SQL JOINS: Exercise-3 with Solution
From the following tables write a SQL query to find the salesperson(s) and the customer(s) he represents. Return Customer Name, city, Salesman, commission.
Sample table: customer
Sample table: salesman
SELECT a.cust_name AS "Customer Name", a.city, b.name AS "Salesman", b.commission FROM customer a INNER JOIN salesman b ON a.salesman_id=b.salesman_id;
Output of the Query:
Customer Name city Salesman commission Nick Rimando New York James Hoog 0.15 Brad Davis New York James Hoog 0.15 Graham Zusi California Nail Knite 0.13 Julian Green London Nail Knite 0.13 Fabian Johnson Paris Mc Lyon 0.14 Geoff Cameron Berlin Lauson Hen 0.12 Jozy Altidor Moscow Paul Adam 0.13 Brad Guzan London Pit Alex 0.11
The said SQL query that is used to select specific columns from two tables, customer and salesman, and join them using the 'salesman_id' column. The query selects the 'cust_name' column from the customer table as 'Customer Name', the 'city' column from the customer table, the 'name' column from the salesman table as 'Salesman', and the 'commission' column from the salesman table.
The query uses an inner join, which only returns rows where there is a match in both tables on the specified join column.
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Previous SQL Exercise: Customers and their cities for given range of orders.
Next SQL Exercise: A salesperson who gets a commission of at least 12%.
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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;
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