SQL Exercises: Salesmen who do not have customers in their cities
SQL UNION : Exercise-6 with Solution
6. From the following tables, write a SQL query to find those salespeople who live in the same city where the customer lives as well as those who do not have customers in their cities by indicating 'NO MATCH'. Sort the result set on 2nd column (i.e. name) in descending order. Return salesperson ID, name, customer name, commission.
Sample table: Salesman
Sample table: Customer
Sample Solution:
SELECT salesman.salesman_id, name, cust_name, commission
FROM salesman, customer
WHERE salesman.city = customer.city
UNION
(SELECT salesman_id, name, 'NO MATCH', commission
FROM salesman
WHERE NOT city = ANY
(SELECT city
FROM customer))
ORDER BY 2 DESC
Sample Output:
salesman_id name cust_name commission 5005 Pit Alex Julian Green 0.11 5005 Pit Alex Brad Guzan 0.11 5007 Paul Adam NO MATCH 0.13 5002 Nail Knite Fabian Johnson 0.13 5006 Mc Lyon Fabian Johnson 0.14 5003 Lauson Hen NO MATCH 0.12 5001 James Hoog Nick Rimando 0.15 5001 James Hoog Brad Davis 0.15
code Explanation:
The said query in SQL that retrieves data from the 'salesman' and 'customer' tables, and outputs a list of salespeople and their commissions based on the city where they are located and the city of their customers.
The query first joins the 'salesman' and 'customer' tables based on the condition that the salesman's city matches the customer's city. It then uses the UNION operator to combine this result with another query that selects salespeople who do not have any customers in any city. For these salespeople, the query outputs "NO MATCH" instead of the customer name.
The ORDER BY clause sorted the output by the second column in descending order.
Practice Online

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Previous SQL Exercise: Largest and smallest orders on each date.
Next SQL Exercise: Any salesman was matched to the city of any customer.
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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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