我要做的是我在数据库中有一个后期/长期存储与他们的条目。我想计算用户lat/long和条目lat/long(在DB中)之间的距离。之后，我想对距离小于500米的进行回声。到目前为止，我能够使用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);
?>
我不建议在SQL语句中转储距离计算，尽管我承认'denil'提供的解决方案非常巧妙。
有3个缺点:代码维护，sql server过载和(最重要的是)地球是不对称的(它就像一个旧的凹陷的棒球被卡车碾过)。这意味着你可能想在将来更改代码(有一些非常复杂的算法在那里- http://en.wikipedia.org/wiki/Geographical_distance)。
我建议使用一个单独的函数，该函数使用一个简单的通用算法(类似于denil的算法)来计算距离。我提交的这段代码是纯PHP(不需要使用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 $geopointDistance;
}
// YOUR CODE HERE
echo distanceGeoPoints(22,50,22.1,50.1);
?>
有许多免费软件(试试GPS trackmaker)，可以让你检查你所在地区的误差范围(如果你需要精度)。对于上述的纬度/长度对，误差在+/- 0.1%以内(根据当地地形学家)。
注意:这个公式给出的是制图距离(海平面距离)，而不是地形距离(不考虑地形)。
我几周前就这么做了。
这个链接是你最好的选择:
http://code.google.com/apis/maps/articles/phpsqlsearch.html即使你不使用他们的API，他们的PHP和SQL查询真的很有帮助。
试试这个查询。我在谷歌上找到了这个，但忘了是谁创建的
SELECT a.*,
3956 * 2 * ASIN(SQRT( POWER(SIN(($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
$lat和$long变量是用户的当前位置。
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 $miles;
}
}
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br>";
?>
测试了3个函数和3个查询，只有一个显示良好的距离:
在米:SELECT *,
(
(((acos(sin((".$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
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
这个查询对我来说是完美的:
$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
最简单的方法
<?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);
?>
使用距离矩阵google api计算经纬度之间的距离
$ch = curl_init();
curl_setopt($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'];
要使用这个API，您将需要key，了解更多关于如何生成key访问https://developers.google.com/maps/documentation/distance-matrix/intro
对于那些试图远离Google(和其他)api的人，我已经使用这个api一段时间了。
因为地球是圆的，对于大尺度半径的提交会有一些奇怪的结果。在相距500英里以内的地点可以很好地工作。
/**
* The max Latitude and Longitude coordinates within a specified milage radius.
*
* @param int $miles
* @param float $longitude
* @param float $latitude
*
* @return array
*/
public function getMaxCoordinates($miles = 50, $longitude, $latitude) {
$oneDegree = 69; // 69 Miles = 1 degree
// Calculate the minimum/maximum possible coordinates.
$lng_min = $longitude - $miles / abs(cos(deg2rad($latitude)) * $oneDegree);
$lng_max = $longitude + $miles / abs(cos(deg2rad($latitude)) * $oneDegree);
$lat_min = $latitude - ($miles / $oneDegree);
$lat_max = $latitude + ($miles / $oneDegree);
return ([
'lat_max' => $lat_max,
'lat_min' => $lat_min,
'lng_max' => $lng_max,
'lng_min' => $lng_min,
'miles' => $miles,
]);
}