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SQL Exercises: Find all customers with orders on October 5, 2012

SQL Query on multiple tables : Exercise-8 with Solution

From the following table, write a SQL query to find those customers who placed orders on October 5, 2012. Return customer_id, cust_name, city, grade, salesman_id, ord_no, purch_amt, ord_date, customer_id and salesman_id.

Sample table: Customer


Sample table: Orders


Sample Solution:

SELECT *
FROM customer a,orders  b 
WHERE a.customer_id=b.customer_id 
AND b.ord_date='2012-10-05';

Output of the Query:

customer_id	cust_name	city		grade	salesman_id	ord_no	purch_amt	ord_date	customer_id	salesman_id
3002		Nick Rimando	New York	100	5001		70002	65.26		2012-10-05	3002		5001
3005		Graham Zusi	California	200	5002		70001	150.50		2012-10-05	3005		5002

Code Explanation:

The said query in SQL that performs a join between two tables: 'customer' and 'orders'. It returns all columns (*) from both tables where the customer_id column in table 'customer' matches the customer_id column in table 'orders' and the ord_date in table 'orders' is equal to '2012-10-05'.

Relational Algebra Expression:

Relational Algebra Expression: Find all customers with orders on October 5, 2012.

Relational Algebra Tree:

Relational Algebra Tree: Find all customers with orders on October 5, 2012.

Explanation:

SQL Subqueries: Find all customers with orders on October 5, 2012.
Inventory database model

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Previous SQL Exercise: Find salesman commission details of given customer.
Next SQL Exercise: SQL Exercises, Practice, Solution - JOINS.

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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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