Postgres:varias uniones hacen que mi consulta devuelva datos incorrectos

Cree SQL complejo paso a paso.

Esto le da los libros que tienen ambas etiquetas requeridas. Solo es tan confiable como la definición de su tabla. La definición de su tabla no debería permitir que un libro tenga la misma etiqueta dos veces. Necesita una restricción ÚNICA en (book_id, tag_id).

SELECT book_id 
FROM books_tags
WHERE books_tags.tag_id IN (716, 101)
GROUP BY book_id
HAVING COUNT(tag_id) = 2

book_id
--
6
3

Puedes usar eso en un JOIN.

SELECT books.id
FROM books
INNER JOIN (
    SELECT book_id 
    FROM books_tags
    WHERE books_tags.tag_id IN (716, 101)
    GROUP BY book_id
    HAVING COUNT(tag_id) = 2) bt ON bt.book_id = books.id

book_id
--
6
3

Unirse a la tabla de votos debería eliminar book_id 6 del resultado. (Sin votos para 6.)

SELECT books.id
FROM books
INNER JOIN (
    SELECT book_id 
    FROM books_tags
    WHERE books_tags.tag_id IN (716, 101)
    GROUP BY book_id
    HAVING COUNT(tag_id) = 2) bt ON bt.book_id = books.id
INNER JOIN books_votes bv ON bv.book_id = books.id

book_id
--
3

Ahora puede agregar la columna de votos a la consulta.

SELECT books.id, bv.vote
FROM books
INNER JOIN (
    SELECT book_id 
        FROM books_tags
    WHERE books_tags.tag_id IN (716, 101)
    GROUP BY book_id
    HAVING COUNT(tag_id) = 2) bt ON bt.book_id = books.id
INNER JOIN books_votes bv ON bv.book_id = books.id

book_id  vote
--
3        1

Finalmente, puedes sumar los votos.

SELECT books.id, SUM(bv.vote) AS total_votes
FROM books
INNER JOIN (
    SELECT book_id 
        FROM books_tags
    WHERE books_tags.tag_id IN (716, 101)
    GROUP BY book_id
    HAVING COUNT(tag_id) = 2) bt ON bt.book_id = books.id
INNER JOIN books_votes bv ON bv.book_id = books.id
GROUP BY books.id;

book_id  total_votes
--
3        1

Su versión no funciona porque devuelve números de identificación de libros incorrectos. La combinación de JOIN en books_votes y la cláusula WHERE no hace lo que esperaba que hiciera.

SELECT books.id AS books_id
FROM books
JOIN books_votes ON books.id = books_votes.book_id
JOIN books_tags ON books.id = books_tags.book_id
WHERE books_tags.tag_id IN (716, 101)
GROUP BY books.id 

books_id
--
3
2

El Libro 2 se incluye no porque tenga ambas etiquetas, sino porque tiene dos votos.

SELECT books.id AS books_id, books_tags.tag_id, books_votes.vote
FROM books
JOIN books_votes ON books.id = books_votes.book_id
JOIN books_tags ON books.id = books_tags.book_id
WHERE books_tags.tag_id IN (716, 101)
ORDER BY books_id, tag_id

book_id  tag_id     vote
--
2        101        1
2        101        1
3        101        1
3        716        1