PHP传递给json_encode之前

I have 2 mysql queries, one to get all the my friends and the other to get the details and a sum of the points for each of my friends.

$get_my_friends_query = mysqli_query($GLOBALS['connect'], "SELECT friend_user_id FROM tbl_friend_link WHERE tbl_friend_link.your_user_id = 1");

while($get_my_friends_result = mysqli_fetch_assoc($get_my_friends_query))
{
   $get_friend_user_details_query = mysqli_query($GLOBALS['connect'], "SELECT tbl_registered_user.user_id, tbl_registered_user.first_name, tbl_registered_user.last_name,  tbl_registered_user.profile_name, SUM(tbl_point.amount) FROM tbl_registered_user JOIN tbl_point ON tbl_registered_user.user_id = tbl_point.user_id WHERE tbl_registered_user.status = 1 AND tbl_registered_user.user_id = '" . $get_my_friends_result['friend_user_id'] ."' GROUP BY tbl_registered_user.user_id ORDER BY SUM(tbl_point.amount)");
   $get_friend_user_details_result = mysqli_fetch_assoc($get_friend_user_details_query);

   $result = array(
      "done" => "true",
      "name" => $get_friend_user_details_result['first_name'] . " " . $get_friend_user_details_result['last_name'],
      "profile_name" => $get_friend_user_details_result['profile_name'],
      "userID" => $get_friend_user_details_result['user_id'],
      "points" => $get_friend_user_details_result['SUM(tbl_point.amount)'],
      "number_friends" => mysqli_num_rows($get_my_friends_query),
      "error_message" => ""
   );
   $row_array[] = $result;
}

echo json_encode($row_array);

I am using the first query because I would like a result to be display regardless of whether they have any points or not. I would like to sort the results based on the SUM of the points however that information is being retrieved in the second query however I could not find a suitable explanation on how to use usort on the $row_array.

I got it to work with the following:

function sorter($key){
    return function ($a, $b) use ($key) {
        return $a[$key] - $b[$key];
    };
}
usort($row_array, sorter('points'));
var_dump($row_array);

usort($row_array, function($a, $b)
{
 return ($a['points']<$b['points']) ? 1 : -1;
});

To "display regardless of whether they have any points or not" you should look into outer joins. Example from Sams Teach Yourself MySQL in 10 Minutes:

mysql>  SELECT c.name, o.order_date
    ->  FROM customers c
    ->  LEFT OUTER JOIN orders o
    ->  ON o.customer_code = c.customer_code
    ->  AND o.order_date BETWEEN '2006-02-01' AND '2006-02-28';
+-------------------------+------------+
| name                    | order_date |
+-------------------------+------------+
| Presidents Incorporated | NULL       |
| Science Corporation     | 2006-02-02 |
| Science Corporation     | 2006-02-05 |  
| Musicians of America    | 2006-02-01 |
| Musicians of America    | 2006-02-02 |
+-------------------------+------------+
5 rows in set (0.00 sec)

In addition I'd use the AS keyword to name the columns what I want directly in the query, instead of having to do what you want there with all the get_friend_user_details_result.

To get exactly what you want in a single query improves speed quite a lot and in many cases also clean up your code quite a bit.