﻿ SQL: Salesmen work for one or more customers or yet to join

SQL Exercise: Salesmen work for one or more customers or yet to join

SQL JOINS: Exercise-14 with Solution

Write a SQL statement to make a list for the salesmen who either work for one or more customers or yet to join any of the customer. The customer, may have placed, either one or more orders on or above order amount 2000 and must have a grade, or he may not have placed any order to the associated supplier.

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 a left outer join with the result of the previous join and the 'orders' table based on customer_id
LEFT OUTER JOIN orders c
ON c.customer_id = a.customer_id
-- Filtering the results based on two conditions (purchase amount greater than or equal to 2000 and grade is not null)
WHERE c.purch_amt >= 2000
AND a.grade IS NOT NULL;
``````

Output of the Query:

```cust_name     |city      |grade|Salesman  |ord_no|ord_date  |purch_amt|
--------------|----------|-----|----------|------|----------|---------|
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 that is joining three tables: customer alias a, salesman alias b, and orders alias c. The query is using a RIGHT OUTER JOIN on salesman and customer tables on the 'salesman_id' column, and a LEFT OUTER JOIN on orders and customer tables on the 'customer_id' column.
The query is then selecting several columns from each table: 'cust_name', 'city', 'grade' from 'customer' table, 'name' aliased as 'Salesman' from salesman table, 'ord_no', 'ord_date', 'purch_amt' from orders table.
The query is also filtering the results based on the condition "c.purch_amt >=2000" and "a.grade IS NOT NULL" which means it will only return the rows where the purchase amount is greater than or equal to 2000 and grade is not null.

Visual Explanation:

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