javascript - 如何获取用户的经纬度，计算两个地点之间的距离？

<code><?php
//以当前用户经纬度为中心,查询5000米内的其他用户
$y = 5000 / 110852; //纬度的范围
$x = 5000 / (111320*cos($lat)); //经度的范围
$sql = '
    select * from user where 
    lat >= ($lat-$y) and lat <= ($lat+$y) and 
    lon >= ($lon-$x) and lon <= ($lon+$x);
';</code>

<code><?php
function distance($lat1, $lon1, $lat2, $lon2) {
    $R = 6371393; //地球平均半径,单位米
    $dlat = deg2rad($lat2-$lat1);
    $dlon = deg2rad($lon2-$lon1);
    $a = pow(sin($dlat/2), 2) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * pow(sin($dlon/2), 2);
    $c = 2 * atan2(sqrt($a), sqrt(1-$a));
    $d = $R * $c;
    return round($d);
}
echo distance(0, 0, -1, 0); // 111202米</code>

<code><?php
$arr = array(
    'win' => array(
        'dis' => 1024,
        'age' => 31
    ),
    'osx' => array(
        'dis' => 512,
        'age' => 15
    ),
    'lin' => array(
        'dis' => 512,
        'age' => 25
    )
);
foreach($arr as $k => $v) {
    $sort['dis'][$k] = $v['dis'];
    $sort['age'][$k] = $v['age'];
}
//先按距离升序排序,如果距离相同,则按年龄降序排序
array_multisort($sort['dis'], SORT_ASC, $sort['age'], SORT_DESC, $arr);
echo json_encode($arr);
//{"lin":{"dis":512,"age":25},"osx":{"dis":512,"age":15},"win":{"dis":1024,"age":31}}</code>