从mysql数据库中选择最近的地理位置最快的方法是什么？
从mysql数据库中选择最近的地理位置最快的方法是什么？
提问于 2012-05-03 21:17:09
我在MySQL数据库中有一个表，其中包含地理坐标和其他有关地点的信息。表中的每一行表示一个地理位置，并具有以下坐标:纬度= 45.05235和经度= 8.02354，这是欧洲的某个地方。
我已经在使用索引了，但是我想加快这个过程，因为这些函数被多次使用。
当然，任何其他解决方案也非常受欢迎。
我做了一个获取最近位置的函数(它正在工作，但速度很慢)：
<?php
//Function for getting nearest destinations:
function nearest_destination($lat1,$lon1,$radius,$type,$maxdistance){
//Determine geo bounds:
$lonlow = $lon1 - rad2deg($maxdistance/6371);
$lonhigh = $lon1 + rad2deg($maxdistance/6371);
$latlow = $lat1 - rad2deg($maxdistance/6371);
$lathigh = $lat1 + rad2deg($maxdistance/6371);
//Database details and connect to database
include(realpath($_SERVER["DOCUMENT_ROOT"]).'/connect_to_db.php');
//Set initial counters to zero
$ii=0;
$i=0;
while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
$shortnamelist[$ii]=$row['shortname'];
$fullnamelist[$ii]=$row['fullname'];
$latitudelist[$ii]=$row['latitude'];
$longitudelist[$ii]=$row['longitude'];
$lon2=$row['longitude'];
$lat2=$row['latitude'];
//Calculate the distance:
$delta_lon = $lon2 - $lon1;
$earth_radius = "6371"; # in km
$distance = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($delta_lon)) ;
$distance = acos($distance);
$distance = $earth_radius*$distance;
$distance = round($distance, 4);
$distancelist[$ii] = $distance;
$ii=$ii+1;
}
//Select position of nearest, and select the destination
if(isset($distancelist)){
$minkey=array_keys($distancelist, min($distancelist));
$minkey=$minkey[0];
$fullname=$fullnamelist[$minkey];
$shortname=$shortnamelist[$minkey];
$latitude=$latitudelist[$minkey];
$longitude=$longitudelist[$minkey];
// remove the big arrays to conserve memory:
unset($fullnamelist);
unset($latitudelist);
unset($longitudelist);
unset($distancelist);
unset($shortnamelist);
}
if(isset($destinid)=='TRUE'){
$nearest_destination = array("shortname" => $shortname, "fullname" => $fullname, "latitude" => $latitude, "longitude" => $longitude, "distancelist" => $distancelisting);}
else $nearest_destination = 0;
mysql_close ();
return $nearest_destination;
}
?>这是在特定半径内选择最近位置的函数(有效但速度很慢)：
<?php
//Function for getting nearest destinations:
function nearest_destination($lat1,$lon1,$radius,$type,$maxdistance){
//Determine geo bounds:
$lonlow = $lon1 - rad2deg($maxdistance/6371);
$lonhigh = $lon1 + rad2deg($maxdistance/6371);
$latlow = $lat1 - rad2deg($maxdistance/6371);
$lathigh = $lat1 + rad2deg($maxdistance/6371);
// Convert from string to number:
$lon1=floatval($lon1);
$lat1=floatval($lat1);
//Database details and connect to database
include(realpath($_SERVER["DOCUMENT_ROOT"]).'/connect_to_database.php'); //Get DB login details
//Select data from destinations table:
$sql="SELECT shortname, fullname, latitude, longitude FROM destinations WHERE type='$type' AND longitude > $lonlow AND longitude < $lonhigh AND latitude > $latlow AND latitude < $lathigh";
$result=mysql_query($sql);
//Set initial counter to zero
$i=0;
while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
$lon2=$row['longitude'];
$lat2=$row['latitude'];
$lon2=floatval($lon2);
$lat2=floatval($lat2);
//Calculate the distance:
$delta_lon = $lon2 - $lon1;
$earth_radius = "6371"; # in km
$distance = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($delta_lon)) ;
$distance = acos($distance);
$distance = $earth_radius*$distance;
$distance = round($distance, 4);
//If distance is smaller than the radius the destination is saved in the array:
if($distance<$radius){
$fullname[$i]=$row['fullname'];
$shortname[$i]=$row['shortname'];
$latitude[$i]=$row['latitude'];
$longitude[$i]=$row['longitude'];
$distancelisting[$i] = $distance;
$i=$i+1;
}
}
if(isset($destinid)=='TRUE'){
$nearest_destination = array("shortname" => $shortname, "fullname" => $fullname, "latitude" => $latitude, "longitude" => $longitude, "distancelist" => $distancelisting);
}else $nearest_destination = 0;
mysql_close ();
return $nearest_destination;
}
?>回答 2
Stack Overflow用户
发布于 2012-05-03 21:22:00
使用mysql gis支持将提高您的速度，因为它是为此而创建的。如果你经常阅读和比较距离，这将是值得使用postgis，这是一个全面支持地理空间数据库。它将让您索引您的点，以进行有效的距离查询。MySQL提供的支持有限，并且依赖于GEOS http://trac.osgeo.org/geos/。
http://forge.mysql.com/wiki/GIS_Functions
http://postgis.refractions.net/
最相关的链接来自Anigel的评论，它给出了你的问题的确切答案Fastest Way to Find Distance Between Two Lat/Long Points
Stack Overflow用户
发布于 2012-05-03 21:25:18
您可以根据需要随意修改：
<?php
$center_lat = $_GET["lat"];
$center_lng = $_GET["lng"];
$radius = $_GET["radius"];
$unit = $_GET["unit"];
$unitConst = $unit == "mi" ? 3959 : 6371;
$sql = sprintf("SELECT Address, City, State, Country, PostalCode, PhoneNumber, Lat, Lng, ($unitConst * acos(cos(radians('%s')) * cos(radians(Lat)) * cos(radians(Lng) - radians('%s')) + sin(radians('%s')) * sin(radians(Lat)))) AS Distance FROM destinations WHERE (Lat != '0' AND Lng !=0) HAVING distance < '%s' ORDER BY distance", mysql_real_escape_string($center_lat), mysql_real_escape_string($center_lng), mysql_real_escape_string($center_lat), mysql_real_escape_string($radius));
?>页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/10432222
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