solutions-2.sql

-- Challenge 1
SELECT 
    a.au_id AS "AUTHOR ID",
    a.au_lname AS "LAST NAME",
    a.au_fname AS "FIRST NAME",
    t.title AS "TITLE",
    p.pub_name AS "PUBLISHER"
FROM titleauthor ta
JOIN authors a ON ta.au_id = a.au_id
JOIN titles t ON ta.title_id = t.title_id
JOIN publishers p ON t.pub_id = p.pub_id;

-- Challenge 2
SELECT 
    a.au_id AS "AUTHOR ID",
    a.au_lname AS "LAST NAME",
    a.au_fname AS "FIRST NAME",
    p.pub_name AS "PUBLISHER",
    COUNT(t.title_id) AS "TITLE COUNT"
FROM titleauthor ta
JOIN authors a ON ta.au_id = a.au_id
JOIN titles t ON ta.title_id = t.title_id
JOIN publishers p ON t.pub_id = p.pub_id
GROUP BY a.au_id, p.pub_name;

-- Challenge 3
SELECT 
    a.au_id AS "AUTHOR ID",
    a.au_lname AS "LAST NAME",
    a.au_fname AS "FIRST NAME",
    SUM(s.qty) AS "TOTAL"
FROM titleauthor ta
JOIN authors a ON ta.au_id = a.au_id
JOIN sales s ON ta.title_id = s.title_id
GROUP BY a.au_id
ORDER BY "TOTAL" DESC
LIMIT 3;

-- Challenge 4
SELECT 
    a.au_id AS "AUTHOR ID",
    a.au_lname AS "LAST NAME",
    a.au_fname AS "FIRST NAME",
    COALESCE(SUM(s.qty), 0) AS "TOTAL"
FROM authors a
LEFT JOIN titleauthor ta ON a.au_id = ta.au_id
LEFT JOIN sales s ON ta.title_id = s.title_id
GROUP BY a.au_id
ORDER BY "TOTAL" DESC;
SELECT 
    a.au_id AS "AUTHOR ID",
    a.au_lname AS "LAST NAME",
    a.au_fname AS "FIRST NAME",
    COALESCE(SUM(s.qty), 0) AS "TOTAL"
FROM authors a
LEFT JOIN titleauthor ta ON a.au_id = ta.au_id
LEFT JOIN sales s ON ta.title_id = s.title_id
GROUP BY a.au_id
ORDER BY "TOTAL" DESC;