What I am trying to do is I have entries in the database which have a lat/long stored with them. I want to calculate the distance between users lat/long and entries lat/long (in DB). After that, I want to echo the ones with distance less than 500 meters. So far I am able to do this using foreach.
<?php
mysql_connect("localhost", "beepbee_kunwarh", "kunwar") or die('MySQL Error.');
mysql_select_db("beepbee_demotest") or die('MySQL Error.');
$Lat = $_REQUEST['Lat'];
$long = $_REQUEST['long'];
$query = mysql_query("SELECT a.*, 3956 * 2 * ASIN(SQRT( POWER(SIN(($Lat - Lat) * pi()/180 / 2), 2) + COS($Lat * pi()/180) * COS(Lat * pi()/180) *POWER(SIN(($long - long) * pi()/180 / 2), 2) )) as distance FROM userResponse GROUP BY beepid HAVING distance <= 500 ORDER by distance ASC;");
$data = array();
while ($row = mysql_fetch_array($query)) {
$data[] = $row;
}
echo json_encode($data);
?>
I did this a few weeks ago.
This link is your best bet:
http://code.google.com/apis/maps/articles/phpsqlsearch.html
Even if you don't use their API, their PHP and SQL query helped really well.
i would not recommend dumping distance calculations in your sql statement, even though i admit that the solution presented by 'denil' is ingenious.
there are 3 downsides: code maintenance, sql server overload AND (above all) the earth is not symmetrical (it is like an old dented baseball that was run over by a truck). this means that you might want to change the code in the future (there are some VERY sophisticated algorithms out there - http://en.wikipedia.org/wiki/Geographical_distance).
i recommend using a separate function that calculates distance with a simple common algorithm (similar if not identical to denil's). i submit this code which is pure php (no need to use googlemaps api):
<?php
function distanceGeoPoints ($lat1, $lng1, $lat2, $lng2) {
$earthRadius = 3958.75;
$dLat = deg2rad($lat2-$lat1);
$dLng = deg2rad($lng2-$lng1);
$a = sin($dLat/2) * sin($dLat/2) +
cos(deg2rad($lat1)) * cos(deg2rad($lat2)) *
sin($dLng/2) * sin($dLng/2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$dist = $earthRadius * $c;
// from miles
$meterConversion = 1609;
$geopointDistance = $dist * $meterConversion;
return $geopointDistance;
}
// YOUR CODE HERE
echo distanceGeoPoints(22,50,22.1,50.1);
?>
there are a number of free softwares (try gps trackmaker) that will allow you to check the margin of error for your part of the globe (if you need precision). for the above lat/long pair, the error is within +/- 0.1% (according to local topographers).
ATTENTION: this formula gives you CARTOGRAPHIC distance (distance at sea level), not TOPOGRAPHIC distance (disconsiders topography).
Try this query. I found this one when googling but forgot who created it
SELECT a.*,
3956 * 2 * ASIN(SQRT( POWER(SIN(($lat - lat) * pi()/180 / 2), 2) + COS($lat * pi()/180) * COS(lat * pi()/180) *
POWER(SIN(($long - longi) * pi()/180 / 2), 2) )) as
distance FROM table
GROUP BY id HAVING distance <= 500 ORDER by distance ASC
$lat and $long variable is the current position of user. lat and longi is the latitude and longitudle of entries
http://www.geodatasource.com/developers/php
<?php
/*::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
/*:: :*/
/*:: This routine calculates the distance between two points (given the :*/
/*:: latitude/longitude of those points). It is being used to calculate :*/
/*:: the distance between two locations using GeoDataSource(TM) Products :*/
/*:: :*/
/*:: Definitions: :*/
/*:: South latitudes are negative, east longitudes are positive :*/
/*:: :*/
/*:: Passed to function: :*/
/*:: lat1, lon1 = Latitude and Longitude of point 1 (in decimal degrees) :*/
/*:: lat2, lon2 = Latitude and Longitude of point 2 (in decimal degrees) :*/
/*:: unit = the unit you desire for results :*/
/*:: where: 'M' is statute miles :*/
/*:: 'K' is kilometers (default) :*/
/*:: 'N' is nautical miles :*/
/*:: Worldwide cities and other features databases with latitude longitude :*/
/*:: are available at http://www.geodatasource.com :*/
/*:: :*/
/*:: For enquiries, please contact sales@geodatasource.com :*/
/*:: :*/
/*:: Official Web site: http://www.geodatasource.com :*/
/*:: :*/
/*:: GeoDataSource.com (C) All Rights Reserved 2014 :*/
/*:: :*/
/*::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
function distance($lat1, $lon1, $lat2, $lon2, $unit) {
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
} else if ($unit == "N") {
return ($miles * 0.8684);
} else {
return $miles;
}
}
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br>";
?>
Tested like 3 functions and 3 queries and only one showed good distance:
In meters:
SELECT *,
(
(((acos(sin((".$latitude."*pi()/180)) * sin((`latitude`*pi()/180))+cos((".$latitude."*pi()/180)) * cos((`latitude`*pi()/180)) * cos(((".$longitude."- `longitude`)*pi()/180))))*180/pi())*60*1.1515*1.609344)
* 1000
) as `distance`
FROM `table`
ORDER BY `distance` ASC
In kilometers:
SELECT *,
(
(((acos(sin((".$latitude."*pi()/180)) * sin((`latitude`*pi()/180))+cos((".$latitude."*pi()/180)) * cos((`latitude`*pi()/180)) * cos(((".$longitude."- `longitude`)*pi()/180))))*180/pi())*60*1.1515*1.609344)
) as `distance`
FROM `table`
ORDER BY `distance` ASC
This query was perfect for me:
$latitude = "23.139422"; //your current lat
$longitude = "-82.382617"; //your current long
SELECT ( 3959 * acos( cos( radians( '.$latitude.' ) ) * cos( radians( latitude ) ) *
cos( radians( longitude ) - radians( '.$longitude.' ) ) + sin( radians( '.$latitude.' )
) * sin( radians( latitude ) ) ) ) AS distance from TABLE
HAVING distance <= 100 ORDER BY distance ASC
Simple and easiest way
<?php
$lat1 = Yourstart_latitude;
$lon1 = Yourstart_longitude;
$lat2 = Yourend_latitude;
$lon2 = Yourend_longitude;
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles= $dist * 60 * 1.1515;
$unit = 'K';
$km = $miles*1.609344;
echo number_format($km,1);
?>
Use distance-matrix google api to calculate distance between latitude and longitude.
$ch = curl_init();
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL,
'https://maps.googleapis.com/maps/api/distancematrix/json?origins='.$prev_add.'&destinations='.$curr_add.'&key=keyyouhavetogenerate'
);
$content = curl_exec($ch);
$array = json_decode($content);
$obj = json_decode($content, TRUE);
$distance = $obj['rows'][0]['elements'][0]['distance']['text'];
to use this api you will need key,to know more about how to generate key visit https://developers.google.com/maps/documentation/distance-matrix/intro
来源：https://stackoverflow.com/questions/7672759/how-to-calculate-distance-from-lat-long-in-php