How to find nearest location by latitude and longitude?

Question

This is the loc_coordinate table structure:

Below is the code, to fetch the nearest places from database and display the place name stored in database itself.

<?php
include("config.php");
$lat = "3.107685";
$lon = "101.7624521";

        $sql="SELECT ((ACOS(SIN($lat * PI() / 180) * SIN(lat * PI() / 180) + COS($lat * PI() / 180) * COS(lat * PI() / 180) * COS(($lon – lon) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS 'distance' FROM loc_coordinate HAVING 'distance'<='10' ORDER BY 'distance' ASC";
        $stmt =$pdo->prepare($sql);
        $stmt->execute();


        while($row = $stmt->fetch())
        {
          echo $row['place'];
        }

?>

The error shown for this:

Fatal error: in C:\wamp\www\mysite\by_coor.php on line 8

PDOException: in C:\wamp\www\mysite\by_coor.php on line 8

echo $sql shows this:

SELECT ((ACOS(SIN(3.107685 * PI() / 180) * SIN(lat * PI() / 180) + COS(3.107685 * PI() / 180) * COS(lat * PI() / 180) * COS((101.7624521 – lon) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS 'distance' FROM loc_coordinate HAVING 'distance'<='10' ORDER BY 'distance' ASC

I'm unsure why I'm getting this error. This is the site I referred to for the SQL query: http://zcentric.com/2010/03/11/calculate-distance-in-mysql-with-latitude-and-longitude/

Answer 1

try this

     SELECT * , (3956 * 2 * ASIN(SQRT( POWER(SIN(( $lat - LatOnTable) *  pi()/180 / 2), 2) +COS( $lat * pi()/180) * COS(LatOnTable * pi()/180) * POWER(SIN(( $long - LongOnTable) * pi()/180 / 2), 2) ))) as distance  
from yourTable  
having  distance <= 10 
order by distance

substitute LatOnTable with the latitude table column name , and longOnTable with you longitude column name in your table .

Answer 2

Here's the SQL statement that finds the closest locations within a radius of 10 miles to the given coordinate. It calculates the distance based on the latitude/longitude of that row and the target latitude/longitude, and then asks for only rows where the distance value is less than and equals to 10, orders the whole query by distance. To search by kilometers instead of miles, replace 3959 with 6371.

SELECT
  id, (
    3959 * acos (
      cos ( radians($lat) )
      * cos( radians( tableLatColName ) )
      * cos( radians( tableLogColName ) - radians($long) )
      + sin ( radians($lat) )
      * sin( radians( tableLatColName ) )
    )
  ) AS distance
FROM table_name
HAVING distance <= 10
ORDER BY distance;

This is using the Google Maps API v3 with a MySQL backend-

https://developers.google.com/maps/solutions/store-locator/clothing-store-locator#findnearsql

I was working on same and found this question so Just wanted to share it.. :)

Answer 3

This works for me:

SELECT restoran.id,restoran.restoran , (6371 * 2 * ASIN(SQRT( POWER(SIN(( -6.9831375276568055 - restoran.lat) *  pi()/180 / 2), 2) +COS( -6.9831375276568055 * pi()/180) * COS(restoran.lat * pi()/180) * POWER(SIN(( 110.40925562381744 - restoran.lng) * pi()/180 / 2), 2) ))) as distance  from restoran having  distance <= 10 order by distance

6371 numbers is for convert to km