如何显示数据库中两个用户之间的相互关系？

Question

I'm trying to show relation between two users. I want to insert only one row to Relations table (see structure below) . But my script is showing that user B is friend of himself - that's not right.

This is a structure of my database (with examples) :

Users
id|email|password|name|pic_url|friend_count
1 |a@.. |aaaaaaaa|A   |http...|1
2 |b@...|bbbbbbbb|B   |http...|0

Relations
id|user_id|friend_id|status(0 and 1)
1 |1      |2        |1(if accepted)

What am I doing wrong? Here's my query:

SELECT Relations.friend_id, Users.name,Users.email,Users.pic_url FROM Relations
INNER JOIN Users
ON Relations.friend_id = Users.id

WHERE  Relations.user_id = $user_id
OR Relations.friend_id = $user_id
AND Relations.status = 1

Answer 1

I recommend to insert two rows into Relations table.

id|user_id|friend_id|status(0 and 1)
1 |1      |2        |1(if accepted)
2 |2      |1        |1(if accepted)

For example, user A send friend request to B. Until B accept this request, the status of both rows would be 0 .

Query for accepting: (user B)

UPDATE Relations SET status=1 WHERE (user_id=$user_id AND friend_id=$friend_id) OR (user_id=$friend_id AND friend_id=$user_id)

Query for selecting:

SELECT Relations.friend_id, Users.name, Users.email, Users.pic_url FROM Relations
    INNER JOIN Users
    ON Relations.friend_id = Users.id

    WHERE Relations.user_id = $user_id
    AND Relations.status = 1

---

Update: If you really want to insert only one row, you would need to execute two queries - the one for the friends user B added and another one for the friends that has added User B to their friend list.

SELECT Relations.friend_id, Users.name, Users.email, Users.pic_url FROM Relations
    INNER JOIN Users
    ON Relations.friend_id = Users.id

    WHERE Relations.user_id = $user_id
    AND Relations.status = 1

SELECT Relations.user_id, Users.name, Users.email, Users.pic_url FROM Relations
    INNER JOIN Users
    ON Relations.user_id = Users.id

    WHERE Relations.friend_id = $user_id
    AND Relations.status = 1

Your problem is that you're selecting friend_id which means that if you execute your query as user B, the friend_id would be 2 - user B. (see your table example)

You should take a look to this question for more information about performance.