我想要做的是我在数据库中有条目存储lat / long。我想计算用户lat / long和条目lat / long(在DB中)之间的距离。在那之后,我想回应距离小于500米的那些。到目前为止,我可以使用foreach
执行此操作。
<?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);
?>
答案 0 :(得分:16)
我不建议在你的sql语句中转储距离计算,即使我承认'denil'提供的解决方案是巧妙的。
有3个缺点:代码维护,sql server重载和(首先)地球不对称(它就像一辆被卡车碾过的旧的凹陷棒球)。这意味着您可能希望将来更改代码(有一些非常复杂的算法 - http://en.wikipedia.org/wiki/Geographical_distance)。
我建议使用一个单独的函数,用一个简单的通用算法计算距离(如果与denil相同,则类似)。我提交这个纯PHP的代码(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 $geopointDistance;
}
// YOUR CODE HERE
echo distanceGeoPoints(22,50,22.1,50.1);
?>
有许多免费软件(试试gps trackmaker)可以让你检查地球部分的误差范围(如果你需要精确度)。对于上述纬度/长对,误差在+/- 0.1%之内(根据当地地形测量师的说法)。
注意:此公式为您提供CARTOGRAPHIC距离(海平面距离),而不是TOPOGRAPHIC距离(disconsiders topography)。
答案 1 :(得分:13)
几个星期前我就这样做了。
此链接是您最好的选择:
http://code.google.com/apis/maps/articles/phpsqlsearch.html
即使您不使用他们的API,他们的PHP和SQL查询也能帮助我们。
答案 2 :(得分:6)
尝试此查询。我在google搜索时找到了这个,但忘记了创建它的人
SELECT a.*,
3956 * 2 * ASIN(SQRT( POWER(SIN(($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。 lat和longi是条目的纬度和长度
答案 3 :(得分:5)
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 $miles;
}
}
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br>";
?>
答案 4 :(得分:1)
测试了3个函数和3个查询,只有一个显示了良好的距离:
以米为单位:
SELECT *,
(
(((acos(sin((".$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
公里:
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
答案 5 :(得分:0)
这个查询非常适合我:
$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
答案 6 :(得分:0)
使用距离矩阵google api计算纬度和经度之间的距离。
$ch = curl_init();
curl_setopt($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'];
要使用此API,您需要密钥,以了解有关如何生成密钥访问的更多信息https://developers.google.com/maps/documentation/distance-matrix/intro
答案 7 :(得分:0)
简单易行的方式
<?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);
?>
答案 8 :(得分:0)
对于那些试图远离Google(和其他)API的人来说,我已经使用了一段时间了。
由于地球是圆形的,因此对于大范围的半径提交会有一些奇怪的结果。彼此相距约500英里以内的位置都可以正常工作。
/**
* 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,
]);
}