我目前有一个MySQL表，其结构如下：

id         name     lon            lat  
-----      -----    -----------    -----------
1          Mark     -76.316528     40.036027
2          John     -95.995102     41.25716
3          Paul     -82.337036     29.645095
4          Dave     -82.337036     29.645095
5          Chris    -76.316528     40.036027

我当前查询数据库以查看用户的位置是否在给定位置的特定英里半径内的方式是通过执行此操作。

function Haversine($lat_from, $lon_from, $lat_to, $lon_to) {
    $radius = 6371000;
    $delta_lat = deg2rad($lat_to-$lat_from);
    $delta_lon = deg2rad($lon_to-$lon_from);

    $a = sin($delta_lat/2) * sin($delta_lat/2) +
        cos(deg2rad($lat_from)) * cos(deg2rad($lat_to)) *
        sin($delta_lon/2) * sin($delta_lon/2);
    $c = 2*atan2(sqrt($a), sqrt(1-$a));

    // Convert the distance from meters to miles
    return ceil(($radius*$c)*0.000621371);
}

// Set the given location to NYC
$my_lon = -73.9844;
$my_lat = 40.7590;

// Query the DB for all of the users
$sql = "SELECT * FROM users";
$result = mysqli_query($con, $sql)or die(mysqli_error($con));
$count = mysqli_num_rows($result);
$i = 0;

while($row = mysqli_fetch_assoc($result)) {
    $lon[$i] = $row['lon'];
    $lat[$i] = $row['lat'];

    $i++;
}

for($i=0;$i<$count;$i++) {
    // Calculate the distance between each user's location and my location
    $distance = Haversine($my_lat, $my_lon, $lat[$i], $lon[$i);

    if($distance < 50) {
        echo "Close enough";
    }
}

这仅适用于表中的几百行。但是，既然我有成千上万的行，那么事实证明检查这么多行非常耗时。我想知道是否有一种方法可以使用Haversine公式仅查询半径50英里以内的行。