mysql query - count words in common between users - mysql

Question

I've a very nice query that selects friends of the current user. user_id = 2 in the example. His friend is user_id = 4.
I want the same query to fetch the number of words user_id 2 has with selected friends. In this case they have word = love, and this is also word they both have, so I want in_common row to say = 1.
Is it possible without changing too much current query?
Should I start from scratch?
SQL FIDDLE

Answer 1

Assuming that both users would have an entry for 'love' in the words_en table then something like this maybe:-
SELECT b.name_surname,
b.avatar,
b.friend_words,
(b.friend_msg_id) AS friend_msg_id,
words_common.words_in_common,
COUNT(m.id) AS unread_msg
FROM
(
SELECT a.name_surname as name_surname,
a.avatar as avatar,
GROUP_CONCAT(DISTINCT w.word ORDER BY w.word ASC) AS friend_words,
(a.friend_id) AS friend_msg_id
FROM
(
SELECT f1.asked_user_id AS friend_id,
f1.created,
u.name_surname,
u.avatar
FROM friends AS f1
INNER JOIN friends AS f2
ON f1.asked_user_id = f2.asker_user_id
AND f1.asker_user_id = f2.asked_user_id
INNER JOIN users AS u ON f1.asked_user_id = u.id
WHERE f1.status = 1 AND f2.status = 1
AND f1.asker_user_id = 2
) a
LEFT JOIN connections c ON c.user_id = a.friend_id
AND c.invisible <> 1 AND c.deleted <> 1
LEFT JOIN words_en w ON c.word_id = w.id
GROUP BY 1
) b
LEFT JOIN messages m ON m.to_user_id = 2
AND m.from_user_id = b.friend_msg_id
AND m.seen = 0
LEFT OUTER JOIN
(
SELECT b.user_id AS friend_id, GROUP_CONCAT(a.word) AS words_in_common
FROM words_en a
INNER JOIN words_en b
ON a.word = b.word
WHERE a.user_id = 2
GROUP BY b.user_id
) words_common
ON b.friend_msg_id = words_common.friend_id
GROUP BY b.name_surname, b.avatar, b.friend_words, b.friend_msg_id
ORDER BY unread_msg DESC
EDIT - modification to use connections table to find common words:-
SELECT b.name_surname,
b.avatar,
b.friend_words,
(b.friend_msg_id) AS friend_msg_id,
words_common.words_in_common,
COUNT(m.id) AS unread_msg
FROM
(
SELECT a.name_surname as name_surname,
a.avatar as avatar,
GROUP_CONCAT(DISTINCT w.word ORDER BY w.word ASC) AS friend_words,
(a.friend_id) AS friend_msg_id
FROM
(
SELECT f1.asked_user_id AS friend_id,
f1.created,
u.name_surname,
u.avatar
FROM friends AS f1
INNER JOIN friends AS f2
ON f1.asked_user_id = f2.asker_user_id
AND f1.asker_user_id = f2.asked_user_id
INNER JOIN users AS u ON f1.asked_user_id = u.id
WHERE f1.status = 1 AND f2.status = 1
AND f1.asker_user_id = 2
) a
LEFT JOIN connections c ON c.user_id = a.friend_id
AND c.invisible <> 1 AND c.deleted <> 1
LEFT JOIN words_en w ON c.word_id = w.id
GROUP BY 1
) b
LEFT JOIN messages m ON m.to_user_id = 2
AND m.from_user_id = b.friend_msg_id
AND m.seen = 0
LEFT OUTER JOIN
(
SELECT b.user_id AS friend_id, GROUP_CONCAT(c.word) AS words_in_common
FROM connections a
INNER JOIN connections b
ON a.word_id = b.word_id
INNER JOIN words_en c
ON b.word_id = c.id
WHERE a.user_id = 2
GROUP BY b.user_id
) words_common
ON b.friend_msg_id = words_common.friend_id
GROUP BY b.name_surname, b.avatar, b.friend_words, b.friend_msg_id
ORDER BY unread_msg DESC