SQL Exercise: Customer who works either through a salesman or by own
SQL JOINS: Exercise-8 with Solution
From the following tables write a SQL query to display the customer name, customer city, grade, salesman, salesman city. The results should be sorted by ascending customer_id.
Sample table: customer
Sample table: salesman
Sample Solution:
SELECT a.cust_name,a.city,a.grade,
b.name AS "Salesman",b.city
FROM customer a
LEFT JOIN salesman b
ON a.salesman_id=b.salesman_id
order by a.customer_id;
Output of the Query:
cust_name city grade Salesman city Brad Guzan London Pit Alex London Nick Rimando New York 100 James Hoog New York Jozy Altidor Moscow 200 Paul Adam Rome Fabian Johnson Paris 300 Mc Lyon Paris Graham Zusi California 200 Nail Knite Paris Brad Davis New York 200 James Hoog New York Julian Green London 300 Nail Knite Paris Geoff Cameron Berlin 100 Lauson Hen San Jose
Explanation:
The said SQL query that uses the LEFT JOIN clause to combine rows from two tables: customer and salesman.
The query retrieves the cust_name, city, grade, name, and city from the two tables where the salesman_id from the customer table matches the salesman_id from the salesman table.
The query also uses the AS keyword to rename the name column as 'Salesman'.
The result of the query will show all rows from the customer table and the matching rows from the salesman table, and any customer without a matching salesman will have NULL values for the salesman columns.
The query also uses the ORDER BY clause to sort the result by customer_id.
Visual Explanation:

Practice Online
Query Visualization:
Duration:

Rows:

Cost:

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Previous SQL Exercise: Join within the tables salesman, customer and orders.
Next SQL Exercise: Customers who holds a grade less than 300.
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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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