# SQL Exercise: One or more customers ordered from the existing list

## SQL JOINS: Exercise-15 with Solution

For those customers from the existing list who put one or more orders, or which orders have been placed by the customer who is not on the list, create a report containing the customer name, city, order number, order date, and purchase amount

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: 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, b.ord_no,
b.ord_date, b.purch_amt AS "Order Amount"
-- Specifying the tables to retrieve data from ('customer' as 'a' and 'orders' as 'b')
FROM customer a
-- Performing a left outer join based on the customer_id, including unmatched rows from 'customer'
LEFT OUTER JOIN orders b
ON a.customer_id = b.customer_id;
``````

Output of the Query:

```cust_name	city		ord_no	ord_date	Order Amount
Brad Guzan	London		70009	2012-09-10	270.65
Nick Rimando	New York	70002	2012-10-05	65.26
Geoff Cameron	Berlin		70004	2012-08-17	110.50
Brad Davis	New York	70005	2012-07-27	2400.60
Nick Rimando	New York	70008	2012-09-10	5760.00
Fabian Johnson	Paris		70010	2012-10-10	1983.43
Geoff Cameron	Berlin		70003	2012-10-10	2480.40
Jozy Altidor	Moscow		70011	2012-08-17	75.29
Nick Rimando	New York	70013	2012-04-25	3045.60
Graham Zusi	California	70001	2012-10-05	150.50
Graham Zusi	California	70007	2012-09-10	948.50
Julian Green	London		70012	2012-06-27	250.45
```

Explanation:

The said SQL query that is joining two tables: customer alias a, and orders alias b. The query is using a LEFT OUTER JOIN on customer and orders tables on the 'customer_id' column.
The query is then selecting several columns from each table: 'cust_name' and 'city' from 'customer' table, 'ord_no', 'ord_date', and 'purch_amt' aliased as 'Order Amount' from orders table.
The LEFT OUTER JOIN ensures that all rows from the customer table are included in the result set, with NULL values in the columns of the orders table for those customers that do not have any matching rows in the orders table.

Visual Explanation:

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Previous SQL Exercise: Salesmen work for one or more customers or yet to join.
Next SQL Exercise: Customer who is neither in the list nor have a grade.

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