Challenge11-FriendBook.sql

/*Challenge: FriendBook */

CREATE TABLE persons (
    id INTEGER PRIMARY KEY AUTOINCREMENT,
    fullname TEXT,
    age INTEGER);
    
INSERT INTO persons (fullname, age) VALUES ("Bobby McBobbyFace", "12");
INSERT INTO persons (fullname, age) VALUES ("Lucy BoBucie", "25");
INSERT INTO persons (fullname, age) VALUES ("Banana FoFanna", "14");
INSERT INTO persons (fullname, age) VALUES ("Shish Kabob", "20");
INSERT INTO persons (fullname, age) VALUES ("Fluffy Sparkles", "8");

CREATE table hobbies (
    id INTEGER PRIMARY KEY AUTOINCREMENT,
    person_id INTEGER,
    name TEXT);
    
INSERT INTO hobbies (person_id, name) VALUES (1, "drawing");
INSERT INTO hobbies (person_id, name) VALUES (1, "coding");
INSERT INTO hobbies (person_id, name) VALUES (2, "dancing");
INSERT INTO hobbies (person_id, name) VALUES (2, "coding");
INSERT INTO hobbies (person_id, name) VALUES (3, "skating");
INSERT INTO hobbies (person_id, name) VALUES (3, "rowing");
INSERT INTO hobbies (person_id, name) VALUES (3, "drawing");
INSERT INTO hobbies (person_id, name) VALUES (4, "coding");
INSERT INTO hobbies (person_id, name) VALUES (4, "dilly-dallying");
INSERT INTO hobbies (person_id, name) VALUES (4, "meowing");

CREATE table friends (
    id INTEGER PRIMARY KEY AUTOINCREMENT,
    person1_id INTEGER,
    person2_id INTEGER);

INSERT INTO friends (person1_id, person2_id)
    VALUES (1, 4);
INSERT INTO friends (person1_id, person2_id)
    VALUES (2, 3);
    
/* In this first step, use a JOIN to display a table showing people's names with their hobbies.*/
	
SELECT persons.fullname, hobbies.name 
FROM persons 
JOIN hobbies 
ON persons.id = hobbies.person_id;

/*Now, use another SELECT with a JOIN to show the names of each pair of friends, based on the data in the friends table.*/

SELECT a.fullname, b.fullname FROM friends
JOIN persons a
ON friends.person1_id = a.id
JOIN persons b
ON friends.person2_id = b.id