MySQLi の「コマンドが同期していない」エラーを解決するにはどうすればよいですか?

$result = mysqli_query($db, $sql) or exit(mysqli_error($db));

if ($result) {
    $result->store_result(); // Buffer the result set
    echo "<table border='1'>
        <tr><th>id</th>
        <th>name</th>
        <th>parent_id</th>
        <th>parent_name</th>
        <th>level</th>
        <th>email</th></tr>";

    while ($row = mysqli_fetch_assoc($result)) {
        $aid = $row["id"];
        $sql2 = "SELECT * FROM members WHERE MEMNO = '$aid'";
        $result2 = mysqli_query($db, $sql2) or exit(mysqli_error($db));

        if ($result = mysqli_fetch_assoc($result2)) {
            $fname = $ newArray['FNAME'];
            $lname = $newArray['LNAME'];
            $mi = $newArray['MI'];
            $address = $newArray['ADDRESS'];
            $city = $newArray['CITY'];
            $state = $newArray['STATE'];
            $zip = $newArray['ZIP'];
            $kdate = $newArray['KDATE'];
            $date = abs(strtotime(date('m/d/Y')) - strtotime(date($kdate))) / (60 * 60 * 24);
        }

        echo sprintf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td></tr>",
        $row["id"],$row["name"],
        $row["parent_id"],$row["parent_name"],
        $row["level"],$row["email"]);
    }

    echo "</table>";
}
mysqli_free_result($result);
mysqli_close($db);