列表建議朋友

Question

我試圖得到的建議朋友說不是我的朋友中，但在我的朋友，朋友的列表。

2個狀態是一個公認的朋友。

我的會話ID是34，我與用戶3和用戶3的朋友與用戶16的朋友，所以用戶16將顯示作爲建議的朋友，因爲用戶16不是我的朋友。

朋友表

id | user 1_id | user2_id | status 
---------------------------------- 
4  3   34  2 
3  34   3   2 
2  3   16  2 
1  16   3   2 

查詢（我試過）

$user1_id=$_SESSION['id']; 
$user2_id=$data['id']; 
$collectmutualfriendsa = mysqli_query($mysqli," 
SELECT DISTINCT r2.user1_id 
FROM 
friends r 
INNER JOIN friends r2 
ON r.user1_id = r2.user2_id 
LEFT OUTER JOIN friends r3 ON r3.user2_id = r2.user1_id AND r3.user1_id=2 
WHERE r.user2_id = 2 AND r3.user1_id is null"); 

Answer 1

好吧，我嘗試了UNION，現在有工作。

$collectmutualfriendsa = mysqli_query($mysqli," 
SELECT * 
FROM friends 
WHERE user2_id NOT IN 
(
    SELECT user1_id FROM friends WHERE user2_id = '$user1_id' 
    UNION 
    SELECT user2_id FROM friends WHERE user2_id = '$user1_id' 
) 

"); 

Answer 2

$user1_id = $_SESSION['id']; // 34 
$user2_id = $data['id']; // 3 
$query = "SELECT * FROM Friends_table WHERE user1_id = '$user2_id' AND user2_id != $user1_id;";