Understanding use of multiple SUMs with LEFT JOINS in mysql - mysql

Question

Using the GROUP BY command, it is possible to LEFT JOIN multiple tables and still get the desired number of rows from the first table.
For example,
SELECT b.title
FROM books `b`
LEFT JOIN orders `o`
ON o.bookid = b.id
LEFT JOIN authors `a`
ON b.authorid = a.id
GROUP BY b.id
However, since behind the scenes MYSQL is doing a cartesian product on the tables, if you include more than one SUM command you get incorrect values based on all the hidden rows. (The problem is explained fairly well here.)
SELECT b.title,SUM(o.id) as sales,SUM(a.id) as authors
FROM books `b`
LEFT JOIN orders `o`
ON o.bookid = b.id
LEFT JOIN authors `a`
ON b.authorid = a.id
GROUP BY b.id
There are a number of answers on SO about this, most using sub-queries in the JOINS but I am having trouble applying them to this fairly simple case.
How can you adjust the above so that you get the correct SUMs?
Edit
Example
books
id|title|authorid
1|Huck Finn|1
2|Tom Sawyer|1
3|Python Cookbook|2
orders
id|bookid
1|1
2|1
3|2
4|2
5|3
6|3
authors
id|author
1|Twain
2|Beazley
2|Jones
The "correct answer" for total # of authors of the Python Cookbook is 2. However, because there are two joins and the overall dataset is expanded by the join on number of orders, SUM(a.id) will be 4.

Answer 1

You are correct that by joining multiple tables you would not get the expected results.
But in this case you should use COUNT() instead of SUM() and count the distinct orders or authors.
Also by your design you should count the names of the authors and not the ids of the table authors:
SELECT b.title,
COUNT(DISTINCT o.id) as sales,
COUNT(DISTINCT a.author) as authors
FROM books `b`
LEFT JOIN orders `o` ON o.bookid = b.id
LEFT JOIN authors `a` ON b.authorid = a.id
GROUP BY b.id, b.title
See the demo.
Results:
| title | sales | authors |
| --------------- | ----- | ------- |
| Huck Finn | 2 | 1 |
| Tom Sawyer | 2 | 1 |
| Python Cookbook | 2 | 2 |

Answer 2

When dealing with separate aggregates, it is good style to aggregate before joining.
Your data model is horribly confusing, making it look like a book is written by one author only (referenced by books.authorid), while this "ID" is not an author's ID at all.
Your main problem is: You don't count! We count with COUNT. But you are mistakenly adding up ID values with SUM.
Here is a proper query, where I am aggregating before joining and using alias names to fight confusion and thus enhance the query's readability and maintainability.
SELECT
b.title,
COALESCE(o.order_count, 0) AS sales,
COALESCE(a.author_count, 0) AS authors
FROM (SELECT title, id AS book_id, authorid AS author_group_id FROM books) b
LEFT JOIN
(
SELECT id as author_group_id, COUNT(*) as author_count
FROM authors
GROUP BY id
) a ON a.author_group_id = b.author_group_id
LEFT JOIN
(
SELECT bookid AS book_id, COUNT(*) as order_count
FROM orders
GROUP BY bookid
) o ON o.book_id = b.book_id
ORDER BY b.title;

Answer 3

i don't think that your query would work like you eexspected.
Assume one book could have 3 authors.
For Authors:
So you would have three rows for that book in your books table,each one for every Author.
So a
SUM(b.authorid)
gives you the correct answer in your case.
For Orders:
you must use a subselect like
LEFT JOIN (SELECT SUM(id) o_sum,bookid FROM orders GROUP BY bookid) `o`
ON o.bookid = b.id
You should really reconsider your approach with books and authors.