使用php mysql获得两个结果

I am creating a friend list using mysql and php. My problem is that only one result is showing up, even though there are two friends for the user i am testing with. my code look as follows:

  <?php 

    if (isset($_POST['tag']) && !empty($_POST['tag'])) {
        $tag = $_POST['tag'];
        require_once 'DB_Functions.php';
        $db = new DB_Functions();
        $response = array("tag" => $tag, "success" => 0, "error" => 0);
        if($tag == 'friend_id_tag') {
            $uuid = $_POST['uuid'];
            if($db->userExists($uuid)) {
                $data = $db->getFriendList($uuid);
                if($data != false) {
                    $response["success"] = 1;   
                    $response["friend"] = $data["name"];
                    echo json_encode($response);
                }else {
                    echo "No existing user";
                }
            } else {
                echo "No existing user";
            }
        }
    }

    ?>

and the mysql query

    <?php

    class DB_Functions {
        private $db;
        //put your code here
        // constructor
        function __construct() {
        require_once 'DB_Connect.php';
        // connecting to database
        $this->db = new DB_Connect();
        $this->db->connect();
        }
    // destructor
    function __destruct() {

        }
    function getFriendList($uuid) {
        $result = mysql_query("SELECT name FROM Account WHERE unique_id IN(SELECT friend_id FROM FriendList WHERE player_id = '$uuid' AND is_friends = 1)");
        $no_of_rows = mysql_num_rows($result);
        if ($no_of_rows > 0) {
            $result = mysql_fetch_array($result);
            return $result;
        } else {
            return false;
        }
    }
    function userExists($uuid) {
        $result = mysql_query("SELECT * FROM Account WHERE unique_id = '$uuid'");
        $no_of_rows = mysql_num_rows($result);
        if ($no_of_rows > 0) {
            // user existed
            return true;
        } else {
            // user not existed
            return false;
        }
        }
    }

    ?>

You should run while loop in function

function getFriendList($uuid) {
        $result = mysql_query("SELECT name FROM Account WHERE unique_id IN(SELECT friend_id FROM FriendList WHERE player_id = '$uuid' AND is_friends = 1)");
        $no_of_rows = mysql_num_rows($result);
        if ($no_of_rows > 0) {
            $d = array();
            while($data = mysql_fetch_array($result)){
                $d[] = $data;
            }
            return $d;
        } else {
            return false;
        }
    }

Your getFriendList returns an array $data = $db->getFriendList($uuid);

but this part is not looped

if($data != false) { $response["success"] = 1;
$response["friend"] = $data["name"]; echo json_encode($response); }else {

So that would just set $response['friend'] to the first user only.

UPDATE: added loop

if($data != false) { $response["success"] = 1; for($i = 0; $i < count($data); $i++) { $response["friend"][$i] = $data[$i]["name"]; } echo json_encode($response); }else {