Hi! Could we please enable some services and cookies to improve your experience and our website?

SQLize | PHPize | SQLtest

Online Sandbox for SQL and PHP: Write, Run, Test, and Share SQL Queries and PHP Code

A A A
Login    Share code      Blog   FAQ
Copy Format Clear
select week(r.rental_date) as week_num, count(*) as weekly_average_rental_count from rental r join customer c on r.customer_id=c.customer_id where year(r.rental_date)=2005 and month(r.rental_date)=08 and r.rental_date between date_add(r.rental_date, interval -1 week) and date_add(r.rental_date, interval 1 week) group by c.customer_id having count(*)=(select avg(count(*)) from rental group by customer_id)

Stuck with a problem? Got Error? Ask AI support!

Copy Clear
Have a question? Ask in our telegram chat!
Contact us:
© 2025 Free Online MySQL 8.0 Sakila (ReadOnly) Playground: Practice & Master SQL Skills | About project | Privacy Policy and Disclaimer
The site powered by: Contabo
PHP versions: PHP 7.4, PHP 8.0, PHP 8.1, PHP 8.2, PHP 8.3, PHP 8.4
RDBMS versions: MySQL 8.0, MySQL 8.0 Sakila (ReadOnly), MySQL 9.3.0, MariaDB 11.4, MariaDB 11.8, MariaDB 11.8 OpenFlights (ReadOnly), MariaDB 10, MariaDB 10 Sakila (ReadOnly), SQLite 3, SQLite 3 Preloaded, PostgreSQL 10 Bookings (ReadOnly), PostgreSQL 14, PostgreSQL 15, PostgreSQL 16, PostgreSQL 17 + PostGIS, PostgreSQL 17 + PostGIS WorkShop (ReadOnly), PostgreSQL 18, MS SQL Server 2017, MS SQL Server 2019, MS SQL Server 2022, MS SQL Server 2022 AdventureWorks (ReadOnly), MS SQL Server 2025, Firebird 4.0, Firebird 4.0 (Employee), Firebird 5.0, RedDatabase 5.0, Oracle Database 19c (HR), Oracle Database 21c, Oracle Database 26ai, SOQOL, ClickHouse