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
customer_id | cust_name | city | grade | salesman_id -------------+----------------+------------+-------+------------- 3002 | Nick Rimando | New York | 100 | 5001 3007 | Brad Davis | New York | 200 | 5001 3005 | Graham Zusi | California | 200 | 5002 3008 | Julian Green | London | 300 | 5002 3004 | Fabian Johnson | Paris | 300 | 5006 3009 | Geoff Cameron | Berlin | 100 | 5003 3003 | Jozy Altidor | Moscow | 200 | 5007 3001 | Brad Guzan | London | | 5005
Sample table: salesman
salesman_id | name | city | commission -------------+------------+----------+------------ 5001 | James Hoog | New York | 0.15 5002 | Nail Knite | Paris | 0.13 5005 | Pit Alex | London | 0.11 5006 | Mc Lyon | Paris | 0.14 5007 | Paul Adam | Rome | 0.13 5003 | Lauson Hen | San Jose | 0.12
Sample table: orders
ord_no purch_amt ord_date customer_id salesman_id ---------- ---------- ---------- ----------- ----------- 70001 150.5 2012-10-05 3005 5002 70009 270.65 2012-09-10 3001 5005 70002 65.26 2012-10-05 3002 5001 70004 110.5 2012-08-17 3009 5003 70007 948.5 2012-09-10 3005 5002 70005 2400.6 2012-07-27 3007 5001 70008 5760 2012-09-10 3002 5001 70010 1983.43 2012-10-10 3004 5006 70003 2480.4 2012-10-10 3009 5003 70012 250.45 2012-06-27 3008 5002 70011 75.29 2012-08-17 3003 5007 70013 3045.6 2012-04-25 3002 5001
Sample Solution:
-- Selecting specific columns and renaming one column for clarity
SELECT a.cust_name, a.city, a.grade,
b.name AS "Salesman",
c.ord_no, c.ord_date, c.purch_amt
-- Specifying the tables to retrieve data from ('customer' as 'a', 'salesman' as 'b', and 'orders' as 'c')
FROM customer a
-- Performing a right outer join based on the salesman_id, including unmatched rows from 'salesman'
RIGHT OUTER JOIN salesman b
ON b.salesman_id = a.salesman_id
-- Performing another right outer join with the result of the previous join and the 'orders' table based on customer_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
Explanation:
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.
Visual Explanation:
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Query Visualization:
Duration:
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