Me sentiría tentado a tener una subconsulta que obtenga todas las palabras que una persona ha aprendido y las una contra sí misma, con las palabras GROUP_CONCAT junto con un conteo. Entonces dando:-
Octopus, NULL, 0
Dog, "Octopus", 1
Spoon, "Octopus,Dog", 2
Así que la subconsulta sería algo como:-
SELECT sub0.idwords, GROUP_CONCAT(sub1.idwords) AS excl_words, COUNT(sub1.idwords) AS older_words_cnt
FROM words_learned sub0
LEFT OUTER JOIN words_learned sub1
ON sub0.userId = sub1.userId
AND sub0.order_learned < sub1.order_learned
WHERE sub0.userId = 1
GROUP BY sub0.idwords
dando
idwords excl_words older_words_cnt
1 NULL 0
2 1 1
3 1,2 2
A continuación, una los resultados de esto con los de otras tablas, buscando artículos en los que coincidan los idwords principales pero no se encuentre ninguno de los otros.
Algo como esto (aunque no probado como sin datos de prueba):-
SELECT sub_words.idwords, words_inc.idArticle
(
SELECT sub0.idwords, SUBSTRING_INDEX(GROUP_CONCAT(sub1.idwords), ',', 10) AS excl_words, COUNT(sub1.idwords) AS older_words_cnt
FROM words_learned sub0
LEFT OUTER JOIN words_learned sub1
ON sub0.userId = sub1.userId
AND sub0.order_learned < sub1.order_learned
WHERE sub0.userId = 1
GROUP BY sub0.idwords
) sub_words
INNER JOIN words words_inc
ON sub_words.idwords = words_inc.idwords
LEFT OUTER JOIN words words_exc
ON words_inc.idArticle = words_exc.idArticle
AND FIND_IN_SET(words_exc.idwords, sub_words.excl_words)
WHERE words_exc.idwords IS NULL
ORDER BY older_words_cnt
LIMIT 100
EDITAR:actualizado para excluir artículos con más de 10 palabras que aún no se han aprendido.
SELECT sub_words.idwords, words_inc.idArticle,
sub2.idArticle, sub2.count, sub2.content
FROM
(
SELECT sub0.idwords, GROUP_CONCAT(sub1.idwords) AS excl_words, COUNT(sub1.idwords) AS older_words_cnt
FROM words_learned sub0
LEFT OUTER JOIN words_learned sub1
ON sub0.userId = sub1.userId
AND sub0.order_learned < sub1.order_learned
WHERE sub0.userId = 1
GROUP BY sub0.idwords
) sub_words
INNER JOIN words words_inc
ON sub_words.idwords = words_inc.idwords
INNER JOIN
(
SELECT a.idArticle, a.count, a.content, SUM(IF(c.idwords_learned IS NULL, 1, 0)) AS unlearned_words_count
FROM Article a
INNER JOIN words b
ON a.idArticle = b.idArticle
LEFT OUTER JOIN words_learned c
ON b.idwords = c.idwords
AND c.userId = 1
GROUP BY a.idArticle, a.count, a.content
HAVING unlearned_words_count < 10
) sub2
ON words_inc.idArticle = sub2.idArticle
LEFT OUTER JOIN words words_exc
ON words_inc.idArticle = words_exc.idArticle
AND FIND_IN_SET(words_exc.idwords, sub_words.excl_words)
WHERE words_exc.idwords IS NULL
ORDER BY older_words_cnt
LIMIT 100
EDITAR:intente comentar la consulta anterior:-
Esto solo selecciona las columnas
SELECT sub_words.idwords, words_inc.idArticle,
sub2.idArticle, sub2.count, sub2.content
FROM
Esta subconsulta obtiene cada una de las palabras aprendidas, junto con una lista separada por comas de las palabras con un order_learned más grande. Esto es para una identificación de usuario en particular
(
SELECT sub0.idwords, GROUP_CONCAT(sub1.idwords) AS excl_words, COUNT(sub1.idwords) AS older_words_cnt
FROM words_learned sub0
LEFT OUTER JOIN words_learned sub1
ON sub0.userId = sub1.userId
AND sub0.order_learned < sub1.order_learned
WHERE sub0.userId = 1
GROUP BY sub0.idwords
) sub_words
Esto es solo para obtener los artículos en los que se usan las palabras (es decir, las palabras aprendidas de la subconsulta anterior)
INNER JOIN words words_inc
ON sub_words.idwords = words_inc.idwords
Esta subconsulta obtiene los artículos que tienen menos de 10 palabras que aún no ha aprendido el usuario en particular.
INNER JOIN
(
SELECT a.idArticle, a.count, a.content, SUM(IF(c.idwords_learned IS NULL, 1, 0)) AS unlearned_words_count
FROM Article a
INNER JOIN words b
ON a.idArticle = b.idArticle
LEFT OUTER JOIN words_learned c
ON b.idwords = c.idwords
AND c.userId = 1
GROUP BY a.idArticle, a.count, a.content
HAVING unlearned_words_count < 10
) sub2
ON words_inc.idArticle = sub2.idArticle
Esta combinación es para encontrar artículos que tienen palabras en la lista separada por comas de la primera subconsulta (es decir, palabras con un orden mayor_aprendidas). Esto se hace como LEFT OUTER JOIN ya que quiero excluir cualquier palabra que se encuentre (esto se hace en la cláusula WHERE al verificar NULL)
LEFT OUTER JOIN words words_exc
ON words_inc.idArticle = words_exc.idArticle
AND FIND_IN_SET(words_exc.idwords, sub_words.excl_words)
WHERE words_exc.idwords IS NULL
ORDER BY older_words_cnt
LIMIT 100
