latitude/longitude find nearest latitude/longitude

Question

I have latitude and longitude and I want to pull the record from the database, which has nearest latitude and longitude by the distance, if that distance gets longer than specified one, then don't retrieve it.

Table structure:

id
latitude
longitude
place name
city
country
state
zip
sealevel

Answer 1

simpledb.execSQL("CREATE TABLE IF NOT EXISTS " + tablename + "(id INTEGER PRIMARY KEY   AUTOINCREMENT,lat double,lng double,address varchar)");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2891001','70.780154','craftbox');");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2901396','70.7782428','kotecha');");//22.2904718 //70.7783906
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2863155','70.772108','kkv Hall');");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.275993','70.778076','nana mava');");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2667148','70.7609386','Govani boys hostal');");


    double curentlat=22.2667258;  //22.2677258
    double curentlong=70.76096826;//70.76096826

    double curentlat1=curentlat+0.0010000;
    double curentlat2=curentlat-0.0010000;

    double curentlong1=curentlong+0.0010000;
    double curentlong2=curentlong-0.0010000;

    try{

        Cursor c=simpledb.rawQuery("select * from '"+tablename+"' where (lat BETWEEN '"+curentlat2+"' and '"+curentlat1+"') or (lng BETWEEN         '"+curentlong2+"' and '"+curentlong1+"')",null);

        Log.d("SQL ", c.toString());
        if(c.getCount()>0)
        {
            while (c.moveToNext())
            {
                double d=c.getDouble(1);
                double d1=c.getDouble(2);

            }
        }
    }
    catch (Exception e)
    {
        e.printStackTrace();
    }

Answer 2

Sounds like you should just use PostGIS, SpatialLite, SQLServer2008, or Oracle Spatial. They can all answer this question for you with spatial SQL.

Answer 3

You're looking for things like the haversine formula. See here as well.

There's other ones but this is the most commonly cited.

If you're looking for something even more robust, you might want to look at your databases GIS capabilities. They're capable of some cool things like telling you whether a point (City) appears within a given polygon (Region, Country, Continent).

Answer 4

Here is my full solution implemented in PHP.

This solution uses the Haversine formula as presented in http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL.

It should be noted that the Haversine formula experiences weaknesses around the poles. This answer shows how to implement the vincenty Great Circle Distance formula to get around this, however I chose to just use Haversine because it's good enough for my purposes.

I'm storing latitude as DECIMAL(10,8) and longitude as DECIMAL(11,8). Hopefully this helps!

showClosest.php

<?PHP
/**
 * Use the Haversine Formula to display the 100 closest matches to $origLat, $origLon
 * Only search the MySQL table $tableName for matches within a 10 mile ($dist) radius.
 */
include("./assets/db/db.php"); // Include database connection function
$db = new database(); // Initiate a new MySQL connection
$tableName = "db.table";
$origLat = 42.1365;
$origLon = -71.7559;
$dist = 10; // This is the maximum distance (in miles) away from $origLat, $origLon in which to search
$query = "SELECT name, latitude, longitude, 3956 * 2 * 
          ASIN(SQRT( POWER(SIN(($origLat - latitude)*pi()/180/2),2)
          +COS($origLat*pi()/180 )*COS(latitude*pi()/180)
          *POWER(SIN(($origLon-longitude)*pi()/180/2),2))) 
          as distance FROM $tableName WHERE 
          longitude between ($origLon-$dist/cos(radians($origLat))*69) 
          and ($origLon+$dist/cos(radians($origLat))*69) 
          and latitude between ($origLat-($dist/69)) 
          and ($origLat+($dist/69)) 
          having distance < $dist ORDER BY distance limit 100"; 
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_assoc($result)) {
    echo $row['name']." > ".$row['distance']."<BR>";
}
mysql_close($db);
?>

./assets/db/db.php

<?PHP
/**
 * Class to initiate a new MySQL connection based on $dbInfo settings found in dbSettings.php
 *
 * @example $db = new database(); // Initiate a new database connection
 * @example mysql_close($db); // close the connection
 */
class database{
    protected $databaseLink;
    function __construct(){
        include "dbSettings.php";
        $this->database = $dbInfo['host'];
        $this->mysql_user = $dbInfo['user'];
        $this->mysql_pass = $dbInfo['pass'];
        $this->openConnection();
        return $this->get_link();
    }
    function openConnection(){
    $this->databaseLink = mysql_connect($this->database, $this->mysql_user, $this->mysql_pass);
    }

    function get_link(){
    return $this->databaseLink;
    }
}
?>

./assets/db/dbSettings.php

<?php
$dbInfo = array(
    'host'      => "localhost",
    'user'      => "root",
    'pass'      => "password"
);
?>

It may be possible to increase performance by using a MySQL stored procedure as suggested by the "Geo-Distance-Search-with-MySQL" article posted above.

I have a database of ~17,000 places and the query execution time is 0.054 seconds.

Answer 5

Try this, it show the nearest points to provided coordinates (within 50 km). It works perfectly:

SELECT m.name,
    m.lat, m.lon,
    p.distance_unit
             * DEGREES(ACOS(COS(RADIANS(p.latpoint))
             * COS(RADIANS(m.lat))
             * COS(RADIANS(p.longpoint) - RADIANS(m.lon))
             + SIN(RADIANS(p.latpoint))
             * SIN(RADIANS(m.lat)))) AS distance_in_km
FROM <table_name> AS m
JOIN (
      SELECT <userLat> AS latpoint, <userLon> AS longpoint,
             50.0 AS radius, 111.045 AS distance_unit
     ) AS p ON 1=1
WHERE m.lat
BETWEEN p.latpoint  - (p.radius / p.distance_unit)
    AND p.latpoint  + (p.radius / p.distance_unit)
    AND m.lon BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
    AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
ORDER BY distance_in_km

Just change <table_name>. <userLat> and <userLon>

You can read more about this solution here: http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

Answer 6

This problem is not very hard at all, but it gets more complicated if you need to optimize it.

What I mean is, do you have 100 locations in your database or 100 million? It makes a big difference.

If the number of locations is small, get them out of SQL and into code by just doing ->

Select * from Location

Once you get them into code, calculate the distance between each lat/lon and your original with the Haversine formula and sort it.