A construction manager works for a home building company. They need to supervise many projects at once.
Users can follow updates on a project
Users can post updates on a project
Users can see 50 most recent updates across projects they follow

Design a relational database schema for this product.

CREATE TABLE IF NOT EXISTS construction.follow_updates (
project_id		         BIGINT(20) NOT NULL AUTO_INCREMENT,
project_name		     VARCHAR(30) NOT NULL,
manager_assigned	     VARCHAR(30) NOT NULL,
comments		         VARCHAR(50), 
project_start_date	DATE		NOT NULL,
		
PRIMARY KEY(project_id)
);
CREATE INDEX project_name ON follow_updates;

CREATE TABLE construction.recent_updates (
project_id_update	BIGINT	NOT NULL	‘Parent foreign key’,
);

select TOP 50 recent_updates.project_id_update, follow_updates.project_id, follow_updates, project_name
from follow_updates
INNER JOIN recent_updates ON 
recent_updates.project_id_update = follow_updates.project_id;