我有个密码：

<?php
require_once('../assets/db/DBConnect.php');
$conn = DBConnect::conn();

echo "<pre>";
$usersWithoutSurnames = [];
$usersWithSurnames = [];
$i = 0;

$query = 'SELECT * FROM person where surName IS NULL';
$res = $conn->query($query);
while($row = $res->fetch_assoc()) {
    if ($i > 7)
        break;
    $usersWithoutSurnames[] = $row;
    $i++;
}

foreach ($usersWithoutSurnames as $user) {
    $getSurnameQuery = 'SELECT clientSecondname from policy WHERE clientEGN = ? ORDER BY id DESC LIMIT 1';

    if($stmt = $conn->prepare($getSurnameQuery)) {
        $egn = $user['egn'];
        $stmt->bind_param("s", $egn);

        $r = $stmt->execute();

        $stmt->bind_result($clientSecondname);
        $stmt->fetch();
        $user['surName'] = $clientSecondname;
        $usersWithSurnames[] = $user;
        $stmt->close();
    }
}

var_dump($usersWithoutSurnames);
echo '-------------------------------------------------';
echo "<pre>"; var_dump($usersWithSurnames); exit;

想法：有一张table users (person) 但有些用户没有姓，所以我查了另一个 table (policy) 对于他们的姓氏，用相同的用户填充第二个数组，但要使用姓氏，然后我将用这个数组填充和表。你可以看到我只得到7个用户，因为当我试图得到更多的用户它超时，我可以优化这个代码/查询与更多的用户一起工作？
表结构：
人：

|------------------------------------------------
|  Id  | firstName | secondName | lastName | EGN
|   1  | test      |   NULL     |  test    | 123
|   2  | test      |   NULL     |  test    | 321

政策：

|------------------------------------------------------------------------
|  Id  | clientFirstname | clientSecondname | clientLastname | clientEGN
|   1  | test            |   test           |  test          |  123
|   2  | test            |   test           |  test          |  321