如何离开与或运算符的连接，并仅确定一个匹配的列

Question

I've search the internet for days and found no solution to my problem. I've two simple tables here:

TB1
tb1_id  tb1_name
1       Red
2       Blue
3       Yellow

TB2
tb2_id  tb2_id1 tb2_id2 tb2_type    tb2_value
1       3       1       A           2

Suppose table 1 is list of users. The guy called Red want to have relationship with a guy called Yellow. So he send a request. When Yellow accepted the request I update tb2_value to 2.

Several days later, Yellow logged in and see who he has relationship with. How to do a query for this where both guys can see each other too. Here's my attempt so far and it's not working:

SELECT
    tb2_id1, tb2_id2, tb2_type, tb2_value, tb1_id, tb1_name
FROM
    TB2
LEFT JOIN
    TB1
    ON
        tb2_id1 = tb1_id OR tb2_id2 = tb1_id
WHERE
    (tb2_id1 = :user_id AND tb2_id2 != :user_id) OR (tb2_id1 != :user_id AND tb2_id2 = :user_id) AND tb2_type = :tb2_type AND tb2_value = :tb2_value AND RAND()<(SELECT ((15/COUNT(*))*10) FROM TB2)
ORDER BY
    RAND()
LIMIT
    15

Or do I do it wrong?

Thank you,

Answer 1

Try this:

SELECT 
    TB2.tb2_id1, TB2.tb2_id2, TB2.tb2_type, TB2.tb2_value,
    IF(TB2.tb2_id1 = :user_id, t2.tb1_id, t1.tb1_id) AS tb1_id, -- :user_id = 1
    IF(TB2.tb2_id1 = :user_id, t2.tb1_name, t1.tb1_name) AS tb1_name -- :user_id = 1
FROM TB2
LEFT JOIN TB1 t1
ON TB2.tb2_id1 = t1.tb1_id
LEFT JOIN TB1 t2
ON TB2.tb2_id2 = t2.tb1_id
WHERE (TB2.tb2_id1 = :user_id OR TB2.tb2_id2 = :user_id) -- :user_id = 1
AND TB2.tb2_type = :tb2_type -- :tb2_type = 'A'
AND TB2.tb2_value = :tb2_value -- :tb2_value = '2'
-- RAND()<(SELECT ((15/COUNT(*))*10) FROM TB2)
ORDER BY RAND()
LIMIT 15

SQLFiddle Demo