Querying MySQL for latitude and longitude coordinates that are within a given mile radius

六眼飞鱼酱① 提交于 2020-01-12 03:35:07

问题


I currently have a MySQL table that is structured as follows:

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

The way I currently query the DB to see if a user's location is within a certain mile radius of a given location is by doing this.

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";
    }
}

This works well with only a few hundred rows in the table. But, now that I have tens of thousands of rows, checking so many rows has proven to be very time consuming. I'm wondering if there is a way to use the Haversine formula to only query rows that are within the 50 mile radius.


回答1:


Spherical Law of Cosines Formula
(37 and -122 are the latitude and longitude of your radius center)

SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) 
    * cos( radians( long ) - radians(-122) ) + sin( radians(37) ) * sin(radians(lat)) ) ) AS distance 
FROM myTable
HAVING distance < 50
ORDER BY distance 

Features

  • Fastest
  • Precision similar to Harvesine Formula

Haversine Formula

SELECT id, 3956 * 2 * ASIN(SQRT(POWER(SIN((37 - abs(lat)) * pi()/180 / 2), 2)
       + COS(37 * pi()/180 ) * COS(abs(lat) * pi()/180)
       * POWER(SIN((-122 - long) * pi()/180 / 2), 2) )) as  distance
FROM myTable
HAVING distance < 50
ORDER BY distance

Features

  • Fast
  • More robust to floating point errors

Note that 3959 is the Earth radius in miles. Earth radius in km: 6371

You can find more information here



来源:https://stackoverflow.com/questions/29553895/querying-mysql-for-latitude-and-longitude-coordinates-that-are-within-a-given-mi

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!