MySQL: Display one random result which a user has not voted on

Question

I need help on displaying one random result in which the current user has not voted on.

Currently my database setup and the last query I have tried can be found on http://sqlfiddle.com/#!2/2f91b/1

Basically I can isolate each individual item using this query:

SELECT a.img_url, a.item_id, a.user_id, a.img_status, b.item_question, c.user_name, c.user_fbid, d.voter_id, count(d.img_id) AS totalVotes
FROM os_photos a 
LEFT JOIN os_items b ON a.item_id = b.item_id
LEFT JOIN os_users c ON a.user_id = c.user_id
LEFT JOIN os_votes d ON a.img_id = d.img_id
GROUP BY a.img_id
ORDER BY RAND()
LIMIT 1

My Problem is: With the SQL knowledge that I have, I am unable to isolate the results to show only the rows in which user #2 has not voted on. I realize the problem is when I use group by, it combines the voter_id and therefore I am unable to check if user #2 has had any input for the item.

Example:

Item #  |  voter_id
1       |      2
1       |      3
2       |      2
3       |      1
3       |      4
4       |      3
4       |      1
5       |      1
5       |      2

With the above sample set, the resulting item should be either item #3, #4 or any other items which have not been voted on.

Your help, advise and knowledge is greatly appreciated.

Answer 1

To get the items that dont exist you need a LEFT JOIN with condition that would otherwise make a positive match, and then add a WHERE clause matching one of the resulting columns to NULL:

SELECT a.img_url, a.item_id, a.user_id, a.img_status, b.item_question, c.user_name,c.user_fbid, d.voter_id, count(d.img_id) AS totalVotes
FROM os_photos a 
LEFT JOIN os_items b ON a.item_id = b.item_id
LEFT JOIN os_users c ON a.user_id = c.user_id
LEFT JOIN os_votes d ON a.img_id = d.img_id
LEFT JOIN os_votes d2 ON a.img_id = d2.img_id AND d2.voter_id=2
WHERE d2.voter_id IS NULL
GROUP BY a.img_id
ORDER BY RAND()
LIMIT 1