地理位置离我的位置最近的位置(纬度,经度)
我想根据我所在的位置显示特定信息.
I want to show specific information depending on where i am.
我有五个信息不同的城市,我想显示我最接近的城市(信息).
I have five cities with different information, and i want to show that city(information) that i'm closest to.
我如何用最简单的方式做到这一点,使用 javascript.
How to i do that the simplest way, using javascript.
例如
如果我将城市经纬度存储在一个数组中
If i store the cities lat, long in an array
var cities = [
['new york', '111111', '222222', 'blablabla']
['boston', '111111', '222222', 'blablabla']
['seattle', '111111', '222222', 'blablabla']
['london', '111111', '222222', 'blablabla']
]
并且以我当前的位置(纬度,经度),我想要离我最近的城市.
And with my current location(lat, long) i want the city that i'm closet to.
推荐答案
这是一个使用 HTML5 地理定位来获取用户位置的基本代码示例.然后它调用 NearestCity()
并计算从该位置到每个城市的距离(公里).我继续使用 Haversine 公式,转而使用更简单的毕达哥拉斯公式和 equirectangular 投影来调整经度线的曲率.
Here is a basic code example using HTML5 geolocation to get the user's position. It then calls NearestCity()
and calculates the distance (km) from the location to each city. I passed on using the Haversine formulae and instead used the simpler Pythagoras formulae and an equirectangular projection to adjust for the curvature in longitude lines.
// 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]);
}
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