How can I put a condition on the way of join? - php

Question

I have this table structure:
// QandA
+----+---------------------+----------------------------------------+------+---------+
| Id | title | content | type | related |
+----+---------------------+----------------------------------------+------+---------+
| 1 | title of question 1 | content of question1 | 0 | 1 |
| 2 | | content of first answer for question1 | 1 | 1 |
| 3 | title of question 2 | content of question2 | 0 | 3 |
| 4 | | content of second answer for question1 | 1 | 1 |
| 5 | | content of first answer for question2 | 1 | 3 |
+----+---------------------+----------------------------------------+------+---------+
type column: 0 means it is a question and 1 means it is a answer.
related column: for question this column is containing the id of itself and for answer this column is containing the id of its question.
Also there is other dependent tables:
// Votes
+----+---------+---------+-------+
| id | post_id | user_id | value |
+----+---------+---------+-------+
| 1 | 1 | 1234 | 1 |
| 2 | 2 | 1234 | -1 |
| 3 | 1 | 4321 | 1 |
+----+---------+---------+-------+
// Favorites
+----+---------+---------+
| id | post_id | user_id |
+----+---------+---------+
| 1 | 1 | 1234 |
| 2 | 1 | 4321 |
+----+---------+---------+
Ok well, This is the main note in my question: Favorites table is only belong to the questions (not answers). Answers can never be favorite (just questions can be)
Also here is my query:
SELECT
p.title, p.content,
vv.value AS cuvv -- cuvv is stand for current_user_vote_value,
CASE WHEN ff.id IS NOT NULL THEN '2' ELSE '3' END AS cuf -- current_user_favorite
(SELECT SUM(v.value) FROM Votes v WHERE p.id = v.post_id) AS total_votes,
(SELECT COUNT(1) FROM Favorites f WHERE p.id = f.post_id) AS total_favorites,
FROM QandA p
LEFT JOIN Votes vv ON p.id = vv.post_id AND vv.user_id = :user_id_1
LEFT JOIN favorites ff ON p.id = ff.post_id AND f.user_id = :user_id_2
WHERE p.related = :id
Note: For cuf, 2 means current user has marked this question as favorite and 3 means he didn't have (in other word, 3 means this question isn't favorite for current user).
Ok, let me pass some parameters to query and execute it: (as an example)
$user_id = 1234;
$id = 1;
$sth->bindValue(":user_id_1", $user_id, PDO::PARAM_INT);
$sth->bindValue(":user_id_2", $user_id, PDO::PARAM_INT);
$sth->bindValue(":id", $id, PDO::PARAM_INT);
$sth->execute();
And here is the output:
-- cuvv is stand for current_user_vote_value
-- cuf is stand for current_user_favorite
+--------------+----------------------+------+-----+-------------+-----------------+
| title | content | cuvv | cuf | total_votes | total_favorites |
+--------------+----------------------+------+-----+-------------+-----------------+
| title of ... | content of que ... | 1 | 2 | 2 | 2 |
| | content of fir ... | -1 | 3 | -1 | 0 |
| | content of sec ... | NULL | 3 | 0 | 0 |
+--------------+----------------------+------+-----+-------------+-----------------+
Ok So, What's my question?
These two columns cuf and total_favorites are just belong to questions (type = 0). But my query doesn't know it. I mean my query calculates the number of total favorites for all rows, and I want to know, how can tell it: calculate cuf and total_favorites only for questions, not both questions and answers?
In other word, I need to put a IF condition to check if p.type = 0 then execute these two lines:
(SELECT COUNT(1) FROM Favorites f WHERE p.id = f.post_id) AS total_favorites,
and
LEFT JOIN favorites ff ON p.id = ff.post_id AND f.user_id = :user_id_2
Otherwise doesn't execute those two lines, because if p.type = 1, then those two lines are waste and useless.
How can I implement that condition and improve that query?

Answer 1

One way you may want to try is to query the favorite and votes table only once in subqueries, and calculate both the user and all values at once.
SELECT
q.title, q.content,
IFNULL(vv.user_val, 0) cuvv, IFNULL(vv.all_val, 0) total_votes,
IFNULL(ff.user_fav, 0) cuf, IFNULL(ff.all_fav, 0) total_favorites
FROM QandA q
LEFT JOIN (
SELECT post_id,
SUM(value) all_val, SUM(CASE WHEN user_id=1234 THEN value END) user_val
FROM votes GROUP BY post_id
) vv
ON vv.post_id = q.id
LEFT JOIN (
SELECT post_id,
COUNT(1) all_fav, COUNT(CASE WHEN user_id=1234 THEN 1 END) user_fav
FROM favorites GROUP BY post_id
) ff
ON q.type=0 AND ff.post_id = q.id
WHERE q.related = 1;
An SQLfiddle to test with.

Answer 2

Try this:
SELECT
p.id, p.type,p.title, p.content,
vv.value AS cuvv,
CASE WHEN ff.id IS NOT NULL THEN '2' ELSE '3' END AS cuf,
(SELECT SUM(v.value) FROM Votes v WHERE p.id = v.post_id) AS total_votes,
(SELECT COUNT(1) FROM Favorites f WHERE p.id = f.post_id) AS total_favorites
FROM QandA p
LEFT JOIN Votes vv ON p.id = vv.post_id AND vv.user_id = '1234'
LEFT JOIN Favorites ff ON p.id = ff.post_id AND ff.user_id = '1234'
WHERE p.related = 1 and p.type=0
union all
SELECT
p.id, p.type,p.title, p.content,
vv.value AS cuvv,
'3' AS cuf,
(SELECT SUM(v.value) FROM Votes v WHERE p.id = v.post_id) AS total_votes,
NULL AS total_favorites
FROM QandA p
LEFT JOIN Votes vv ON p.id = vv.post_id AND vv.user_id = '1234'
WHERE p.related = 1 and p.type=1;