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

Sample table: orders

Sample Solution:

SELECT a.cust_name,a.city, b.ord_no,
b.ord_date,b.purch_amt AS "Order Amount" 
FROM customer a 
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


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:

Result of a report with customer name, city, order no. order date, purchase amount for those customers from the existing list who placed one or more orders or which order(s) have been placed by the customer who is not in the list

Practice Online

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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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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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