【发布时间】：2011-04-24 18:11:08
【问题描述】：
我正在制作一个脚本，其中将大量业务加载到具有纬度和经度的 mySQL 数据库中。然后我为该脚本提供纬度和经度（最终用户的），脚本必须计算从提供的纬度/经度到它从数据库中获取的每个条目的距离，并按照最近到最远的顺序对它们进行排序.
我实际上只需要大约 10 或 20 个“最近”结果，但除了从数据库中获取所有结果并在每个结果上运行函数然后进行数组排序之外，我想不出其他办法。
这是我已经拥有的：
<?php
function getDistance($point1, $point2){
$radius = 3958; // Earth's radius (miles)
$pi = 3.1415926;
$deg_per_rad = 57.29578; // Number of degrees/radian (for conversion)
$distance = ($radius * $pi * sqrt(
($point1['lat'] - $point2['lat'])
* ($point1['lat'] - $point2['lat'])
+ cos($point1['lat'] / $deg_per_rad) // Convert these to
* cos($point2['lat'] / $deg_per_rad) // radians for cos()
* ($point1['long'] - $point2['long'])
* ($point1['long'] - $point2['long'])
) / 180);
$distance = round($distance,1);
return $distance; // Returned using the units used for $radius.
}
include("../includes/application_top.php");
$lat = (is_numeric($_GET['lat'])) ? $_GET['lat'] : 0;
$long = (is_numeric($_GET['long'])) ? $_GET['long'] : 0;
$startPoint = array("lat"=>$lat,"long"=>$long);
$sql = "SELECT * FROM mellow_listings WHERE active=1";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
$thedistance = getDistance($startPoint,array("lat"=>$row['lat'],"long"=>$row['long']));
$data[] = array('id' => $row['id'],
'name' => $row['name'],
'description' => $row['description'],
'lat' => $row['lat'],
'long' => $row['long'],
'address1' => $row['address1'],
'address2' => $row['address2'],
'county' => $row['county'],
'postcode' => strtoupper($row['postcode']),
'phone' => $row['phone'],
'email' => $row['email'],
'web' => $row['web'],
'distance' => $thedistance);
}
// integrate google local search
$url = "http://ajax.googleapis.com/ajax/services/search/local?";
$url .= "q=Off+licence"; // query
$url .= "&v=1.0"; // version number
$url .= "&rsz=8"; // number of results
$url .= "&key=ABQIAAAAtG"
."Pcon1WB3b0oiqER"
."FZ-TRQgsWYVg721Z"
."IDPMPlc4-CwM9Xt"
."FBSTZxHDVqCffQ2"
."W6Lr4bm1_zXeYoQ"; // api key
$url .= "&sll=".$lat.",".$long;
// sendRequest
// note how referer is set manually
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_REFERER, /* url */);
$body = curl_exec($ch);
curl_close($ch);
// now, process the JSON string
$json = json_decode($body, true);
foreach($json['responseData']['results'] as $array){
$thedistance = getDistance($startPoint,array("lat"=>$array['lat'],"long"=>$array['lng']));
$data[] = array('id' => '999',
'name' => $array['title'],
'description' => '',
'lat' => $array['lat'],
'long' => $array['lng'],
'address1' => $array['streetAddress'],
'address2' => $array['city'],
'county' => $array['region'],
'postcode' => '',
'phone' => $array['phoneNumbers'][0],
'email' => '',
'web' => $array['url'],
'distance' => $thedistance);
}
// sort the array
foreach ($data as $key => $row) {
$id[$key] = $row['id'];
$distance[$key] = $row['distance'];
}
array_multisort($distance, SORT_ASC, $data);
header("Content-type: text/xml");
echo '<?xml version="1.0" encoding="UTF-8"?>'."\n";
echo '<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">'."\n";
echo '<plist version="1.0">'."\n";
echo '<array>'."\n";
for($i = 0; isset($distance[$i]); $i++){
//echo $data[$i]['id']." -> ".$distance[$i]."<br />";
echo '<dict>'."\n";
foreach($data[$i] as $key => $val){
echo '<key><![CDATA['.$key.']]></key>'."\n";
echo '<string><![CDATA['.htmlspecialchars_decode($val, ENT_QUOTES).']]></string>'."\n";
}
echo '</dict>'."\n";
}
echo '</array>'."\n";
echo '</plist>'."\n";
?>
现在，在数据库中只有 2 或 3 个企业的情况下，它运行得足够快，但我目前正在将 5k 企业加载到数据库中，我担心每个条目运行它会非常慢？你怎么看？
它也不是我可以缓存的那种数据，因为两个用户具有相同纬度/经度的可能性非常罕见，因此无济于事。
对此我能做些什么？
感谢您的帮助和任何建议。他们都非常感谢。
【问题讨论】：
-
也许你应该改写你的标题，因为你正在寻找一种解决方案，如何有效地找到给定位置附近的最近位置。