SQL Exercise: Customers from a different city than the salesmen
SQL JOINS: Exercise-20 with Solution
Write a SQL statement to make a Cartesian product between salesman and customer i.e. each salesman will appear for all customers and vice versa for those salesmen who must belong to a city which is not the same as his customer and the customers should have their own grade.
Sample table: salesman
Sample table: customer
Sample Solution:
SELECT *
FROM salesman a
CROSS JOIN customer b
WHERE a.city IS NOT NULL
AND b.grade IS NOT NULL
AND a.city<>b.city;
Output of the Query:
salesman_id name city commission customer_id cust_name city grade salesman_id 5002 Nail Knite Paris 0.13 3002 Nick Rimando New York 100 5001 5005 Pit Alex London 0.11 3002 Nick Rimando New York 100 5001 5006 Mc Lyon Paris 0.14 3002 Nick Rimando New York 100 5001 5007 Paul Adam Rome 0.13 3002 Nick Rimando New York 100 5001 5003 Lauson Hen San Jose0.12 3002 Nick Rimando New York 100 5001 5002 Nail Knite Paris 0.13 3007 Brad Davis New York 200 5001 5005 Pit Alex London 0.11 3007 Brad Davis New York 200 5001 5006 Mc Lyon Paris 0.14 3007 Brad Davis New York 200 5001 5007 Paul Adam Rome 0.13 3007 Brad Davis New York 200 5001 5003 Lauson Hen San Jose0.12 3007 Brad Davis New York 200 5001 5001 James Hoog New York0.15 3005 Graham Zusi California 200 5002 5002 Nail Knite Paris 0.13 3005 Graham Zusi California 200 5002 5005 Pit Alex London 0.11 3005 Graham Zusi California 200 5002 5006 Mc Lyon Paris 0.14 3005 Graham Zusi California 200 5002 5007 Paul Adam Rome 0.13 3005 Graham Zusi California 200 5002 5003 Lauson Hen San Jose0.12 3005 Graham Zusi California 200 5002 5001 James Hoog New York0.15 3008 Julian Green London 300 5002 5002 Nail Knite Paris 0.13 3008 Julian Green London 300 5002 5006 Mc Lyon Paris 0.14 3008 Julian Green London 300 5002 5007 Paul Adam Rome 0.13 3008 Julian Green London 300 5002 5003 Lauson Hen San Jose0.12 3008 Julian Green London 300 5002 5001 James Hoog New York0.15 3004 Fabian Johnson Paris 300 5006 5005 Pit Alex London 0.11 3004 Fabian Johnson Paris 300 5006 5007 Paul Adam Rome 0.13 3004 Fabian Johnson Paris 300 5006 5003 Lauson Hen San Jose0.12 3004 Fabian Johnson Paris 300 5006 5001 James Hoog New York0.15 3009 Geoff Cameron Berlin 100 5003 5002 Nail Knite Paris 0.13 3009 Geoff Cameron Berlin 100 5003 5005 Pit Alex London 0.11 3009 Geoff Cameron Berlin 100 5003 5006 Mc Lyon Paris 0.14 3009 Geoff Cameron Berlin 100 5003 5007 Paul Adam Rome 0.13 3009 Geoff Cameron Berlin 100 5003 5003 Lauson Hen San Jose0.12 3009 Geoff Cameron Berlin 100 5003 5001 James Hoog New York0.15 3003 Jozy Altidor Moscow 200 5007 5002 Nail Knite Paris 0.13 3003 Jozy Altidor Moscow 200 5007 5005 Pit Alex London 0.11 3003 Jozy Altidor Moscow 200 5007 5006 Mc Lyon Paris 0.14 3003 Jozy Altidor Moscow 200 5007 5007 Paul Adam Rome 0.13 3003 Jozy Altidor Moscow 200 5007 5003 Lauson Hen San Jose0.12 3003 Jozy Altidor Moscow 200 5007
Explanation:
This SQL query is selecting all columns from two tables, salesman and customer, and joining them using a cross join. The query also has several conditions in the WHERE clause:
The city column in the salesman table is not null
The grade column in the customer table is not null
The city column in the salesman table is not equal to the city column in the customer table.
This query is likely returning a large set of data containing all combinations of rows from the two tables, filtered by the conditions in the WHERE clause. The result set will contain all rows where the salesman and customer are from different city.
Visual Explanation:
Practice Online
Query Visualization:
Duration:
Rows:
Cost:
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Previous SQL Exercise: Salesmen belongs to a city, customers have a grade
Next SQL Exercise: Display each item producer company.
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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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