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SQL Exercise: Customer have placed no order or one or more orders

SQL JOINS: Exercise-10 with Solution

Write a SQL statement to make a report with customer name, city, order number, order date, and order amount in ascending order according to the order date to determine whether any of the existing customers have placed an order or not.

Sample table: orders


Sample table: customer


Sample Solution:

SELECT a.cust_name,a.city, b.ord_no,
b.ord_date,b.purch_amt AS "Order Amount" 
FROM customer a 
LEFT OUTER JOIN orders b 
ON a.customer_id=b.customer_id 
order by b.ord_date;

Output of the Query:

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

Explanation:

The said SQL query is selecting the customer name, city from the customer table aliased as a and the order number, order date and purchase amount from the orders table aliased as b. It is joining these tables on the 'customer_id' column, and the results are ordered by the 'ord_date' column.
Additionally, it is using a LEFT OUTER JOIN which will retrieve all records from the left table and the matching records from the right table. If no match is found on the right table, it will return NULL for the right table's fields.

Visual Explanation:

Result of a report with customer name, city, order number, date and amount in ascending order on order date

Practice Online


Query Visualization:

Duration:

Query visualization of Make a report with customer name, city, order number, date, and amount in ascending order on order date to find either any of the existing customer have placed no order or placed one or more orders - Duration

Rows:

Query visualization of Make a report with customer name, city, order number, date, and amount in ascending order on order date to find either any of the existing customer have placed no order or placed one or more orders - Rows

Cost:

Query visualization of Make a report with customer name, city, order number, date, and amount in ascending order on order date to find either any of the existing customer have placed no order or placed one or more orders - Cost

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Previous SQL Exercise: Customers who holds a grade less than 300.
Next SQL Exercise: Find customers who have placed no order or one or more.

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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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