我想找到我的朋友,从我当前所在位置靠近我的手机拍摄他们的位置。 例如,在我下面的代码中,我有各种各样的城市,如果我把3,4个朋友的数量放在一起,那么我可以这样做吗?或者我可以做其他一些改变吗?有可能吗?
// Get User's Coordinate from their Browser
window.onload = function () {
// HTML5/W3C Geolocation
if (navigator.geolocation) {
navigator.geolocation.getCurrentPosition(UserLocation);
}
// Default to Washington, DC
else
NearestCity(38.8951, -77.0367);
}
// Callback function for asynchronous call to HTML5 geolocation
function UserLocation(position) {
NearestCity(position.coords.latitude, position.coords.longitude);
}
// Convert Degress to Radians
function Deg2Rad(deg) {
return deg * Math.PI / 180;
}
function PythagorasEquirectangular(lat1, lon1, lat2, lon2) {
lat1 = Deg2Rad(lat1);
lat2 = Deg2Rad(lat2);
lon1 = Deg2Rad(lon1);
lon2 = Deg2Rad(lon2);
var R = 6371; // km
var x = (lon2 - lon1) * Math.cos((lat1 + lat2) / 2);
var y = (lat2 - lat1);
var d = Math.sqrt(x * x + y * y) * R;
return d;
}
var lat = 20; // user's latitude
var lon = 40; // user's longitude
var cities = [
["city1", 10, 50, "blah"],
["city2", 40, 60, "blah"],
["city3", 25, 10, "blah"],
["city4", 5, 80, "blah"]
];
function NearestCity(latitude, longitude) {
var mindif = 99999;
var closest;
for (index = 0; index < cities.length; ++index) {
var dif = PythagorasEquirectangular(latitude, longitude, cities[index]
[1], cities[index][2]);
if (dif < mindif) {
closest = index;
mindif = dif;
}
}
// echo the nearest city
alert(cities[closest]);
}