SQL Exercise: Salesmen who works either for one or more customer
SQL JOINS: Exercise-13 with Solution
From the following tables write a SQL query to list all salespersons along with customer name, city, grade, order number, date, and amount. Condition for selecting list of salesmen : 1. Salesmen who works for one or more customer or, 2. Salesmen who not yet join under any customer, Condition for selecting list of customer : 3. placed one or more orders, or 4. no order placed to their salesman.
Sample table: customer
Sample table: salesman
Sample table: orders
SELECT a.cust_name,a.city,a.grade, b.name AS "Salesman", c.ord_no, c.ord_date, c.purch_amt FROM customer a RIGHT OUTER JOIN salesman b ON b.salesman_id=a.salesman_id RIGHT OUTER JOIN orders c ON c.customer_id=a.customer_id;
Output of the Query:
cust_name city grade Salesman ord_no ord_date purch_amt Brad Guzan London Pit Alex 70009 2012-09-10 270.65 Nick Rimando New York 100 James Hoog 70002 2012-10-05 65.26 Geoff Cameron Berlin 100 Lauson Hen 70004 2012-08-17 110.50 Brad Davis New York 200 James Hoog 70005 2012-07-27 2400.60 Nick Rimando New York 100 James Hoog 70008 2012-09-10 5760.00 Fabian Johnson Paris 300 Mc Lyon 70010 2012-10-10 1983.43 Geoff Cameron Berlin 100 Lauson Hen 70003 2012-10-10 2480.40 Jozy Altidor Moscow 200 Paul Adam 70011 2012-08-17 75.29 Nick Rimando New York 100 James Hoog 70013 2012-04-25 3045.60 Graham Zusi California 200 Nail Knite 70001 2012-10-05 150.50 Graham Zusi California 200 Nail Knite 70007 2012-09-10 948.50 Julian Green London 300 Nail Knite 70012 2012-06-27 250.45
The said SQL query is performing a right outer join on three tables: the customer table alias a, the salesman table alias b, and the orders table alias c.
First, it performs a right outer join on the customer table and the salesman table using the 'salesman_id' column.
Then it performs another right outer join on the result of the first join and the orders table using the 'customer_id' column.
It is then selecting the 'cust_name', 'city', 'grade', 'name' as 'Salesman', 'ord_no', 'ord_date', and 'purch_amt' columns from the three tables.
This query will select all the rows from orders table and any matching rows from customer table and salesman table. If there is no match, it will return NULL for the non-matching columns of customer table and salesman table.
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous SQL Exercise: Salesmen works either for one or more customer or not.
Next SQL Exercise: Salesmen work for one or more customers or yet to join.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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;
- Weekly Trends
- Python Interview Questions and Answers: Comprehensive Guide
- Scala Exercises, Practice, Solution
- Kotlin Exercises practice with solution
- MongoDB Exercises, Practice, Solution
- SQL Exercises, Practice, Solution - JOINS
- Java Basic Programming Exercises
- SQL Subqueries
- Adventureworks Database Exercises
- C# Sharp Basic Exercises
- SQL COUNT() with distinct
- Java Collection Exercises
- SQL COUNT() function
- SQL Inner Join
We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook