Question

我试图创建一个共同的朋友查询。但我不确定如何合并两个表以获得正确的结果。

因此，如果会话用户与他们的朋友有共同的朋友，那么共同的朋友就是其中之一。我正在寻找的是最好的解决方案，最快和最小的查询来完成这项工作。

用户表

id | first | last | username
----------------------------

朋友表

id | user 1_id | user2_id | status
----------------------------------

尝试查询

 $collectmutualfriends = mysqli_query($mysqli,"
    SELECT 
               user.id,
               user.first,
               user.last,
               user.username
    FROM       user 
    LEFT JOIN friends as friends1
           ON  friends1.user1_id = user.id
           AND friends1.user2_id = '$user2_id'
           AND friends1.status = 2

    LEFT JOIN friends as friends2
           ON  friends2.user2_id = user.id
           AND friends2.user1_id = '$user2_id'
           AND friends2.status = 2

    GROUP BY friends2.user2_id
    ");
$nummutual = mysqli_num_rows($collectmutualfriends);
if($nummutual==1){
    echo"<a href='mutualfriends.php?username=".$_GET['username']."'>".$nummutual." Mutual Friend<br></a>";
    }else if($nummutual>1){
    echo"<a href='mutualfriends.php?username=".$_GET['username']."'>".$nummutual." Mutual Friends<br></a>";
    }

更新

$user1_id = $_SESSION['id'];
$user2_id = $data['id'];

$ user1_id和$ user2_id可以在friends表中的user1_id或user2_id中，具体取决于谁先发送好友请求。

Answer 1

这应该可以解决问题：

$sql = "SELECT u.*
          FROM friends f1 
    INNER JOIN friends f2 ON (f2.user2_id = f1.user2_id)
    INNER JOIN user u ON (u.id = f2.user2_id)
         WHERE f1.user1_id = '$user1_id'
           AND f2.user1_id = '$user2_id'";

相互朋友查询

1 个答案: