如何检查坐标是否在公里半径内?
How to check if coords are within a radius in km?
给定 json
与 坐标:
var centerLat = 51.6000;
var centerLng = 12.8000;
var posts = [
{
name: 'PostA',
latitude: '52.5167',
longitude: '13.3833',
},
{
name: 'PostB',
latitude: '52.9667',
longitude: '13.7167',
},
{
name: 'PostC',
latitude: '26.7767',
longitude: '18.4567',
}
];
我在 this link 上找到了 haversine 公式,我如何检查给定 lat
和 lng
的列表是否来自json
在 5 km 的 radius
范围内,使用 haversine?
function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2-lat1); // deg2rad below
var dLon = deg2rad(lon2-lon1);
var a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180)
}
你可以这样做来过滤城市,你提供的数据相距太远,为了在结果中得到任何东西我已经移动了 centerLat
和 centerLng
我已经记录了数组最近的城市在最后。
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2 - lat1); // deg2rad below
var dLon = deg2rad(lon2 - lon1);
var a =
Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI / 180)
}
var centerLat = 52.5167;
var centerLng = 13.3933;
var posts = [{
name: 'PostA',
latitude: '52.5167',
longitude: '13.3833',
},
{
name: 'PostB',
latitude: '52.9667',
longitude: '13.7167',
},
{
name: 'PostC',
latitude: '26.7767',
longitude: '18.4567',
}
];
let closePosts = [];
posts.forEach((post) => {
if (getDistanceFromLatLonInKm(centerLat, centerLng, post.latitude, post.longitude) < 5) {
closePosts.push(post);
}
});
console.log(closePosts)
只需将此添加到您的代码中就足够了。
posts.forEach(post => {
const distance = getDistanceFromLatLonInKm(centerLat, centerLng, post.latitude, post.longitude);
if (distance <= 200) {
console.log(`Distance to ${post.name}: ${distance} km`);
}
});
但我建议多清理一下代码。我使用 200 而不是 5,因为您的示例没有产生结果。
这是一个重构的片段。
const posts = [
{
name: 'Berlin',
latitude: '52.520008',
longitude: '13.404954',
},
{
name: 'Hamburg',
latitude: '53.551086',
longitude: '9.993682',
},
{
name: 'München',
latitude: '48.135124',
longitude: '11.581981',
},
{
name: 'Lübeck',
latitude: '53.865467',
longitude: '10.686559',
},
{
name: 'Schwerin',
latitude: '53.635502',
longitude: '11.401250',
},
];
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
const R = 6371; // Radius of the earth in km
const dLat = deg2rad(lat2-lat1); // deg2rad below
const dLon = deg2rad(lon2-lon1);
const a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
const d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180);
}
function findClosePosts(location, radius, posts) {
return posts.filter((post) =>
// find close points within the radius of the location, but exclude the location itself from results
getDistanceFromLatLonInKm(location.latitude, location.longitude, post.latitude, post.longitude) <= radius && location !== post);
}
function findLocationByName(name, posts) {
return posts.find((post) => post.name === name);
}
const hamburg = findLocationByName('Hamburg', posts);
const closePosts = findClosePosts(hamburg, 200, posts);
console.log(closePosts);
给定 json
与 坐标:
var centerLat = 51.6000;
var centerLng = 12.8000;
var posts = [
{
name: 'PostA',
latitude: '52.5167',
longitude: '13.3833',
},
{
name: 'PostB',
latitude: '52.9667',
longitude: '13.7167',
},
{
name: 'PostC',
latitude: '26.7767',
longitude: '18.4567',
}
];
我在 this link 上找到了 haversine 公式,我如何检查给定 lat
和 lng
的列表是否来自json
在 5 km 的 radius
范围内,使用 haversine?
function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2-lat1); // deg2rad below
var dLon = deg2rad(lon2-lon1);
var a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180)
}
你可以这样做来过滤城市,你提供的数据相距太远,为了在结果中得到任何东西我已经移动了 centerLat
和 centerLng
我已经记录了数组最近的城市在最后。
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2 - lat1); // deg2rad below
var dLon = deg2rad(lon2 - lon1);
var a =
Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI / 180)
}
var centerLat = 52.5167;
var centerLng = 13.3933;
var posts = [{
name: 'PostA',
latitude: '52.5167',
longitude: '13.3833',
},
{
name: 'PostB',
latitude: '52.9667',
longitude: '13.7167',
},
{
name: 'PostC',
latitude: '26.7767',
longitude: '18.4567',
}
];
let closePosts = [];
posts.forEach((post) => {
if (getDistanceFromLatLonInKm(centerLat, centerLng, post.latitude, post.longitude) < 5) {
closePosts.push(post);
}
});
console.log(closePosts)
只需将此添加到您的代码中就足够了。
posts.forEach(post => {
const distance = getDistanceFromLatLonInKm(centerLat, centerLng, post.latitude, post.longitude);
if (distance <= 200) {
console.log(`Distance to ${post.name}: ${distance} km`);
}
});
但我建议多清理一下代码。我使用 200 而不是 5,因为您的示例没有产生结果。
这是一个重构的片段。
const posts = [
{
name: 'Berlin',
latitude: '52.520008',
longitude: '13.404954',
},
{
name: 'Hamburg',
latitude: '53.551086',
longitude: '9.993682',
},
{
name: 'München',
latitude: '48.135124',
longitude: '11.581981',
},
{
name: 'Lübeck',
latitude: '53.865467',
longitude: '10.686559',
},
{
name: 'Schwerin',
latitude: '53.635502',
longitude: '11.401250',
},
];
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
const R = 6371; // Radius of the earth in km
const dLat = deg2rad(lat2-lat1); // deg2rad below
const dLon = deg2rad(lon2-lon1);
const a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
const d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180);
}
function findClosePosts(location, radius, posts) {
return posts.filter((post) =>
// find close points within the radius of the location, but exclude the location itself from results
getDistanceFromLatLonInKm(location.latitude, location.longitude, post.latitude, post.longitude) <= radius && location !== post);
}
function findLocationByName(name, posts) {
return posts.find((post) => post.name === name);
}
const hamburg = findLocationByName('Hamburg', posts);
const closePosts = findClosePosts(hamburg, 200, posts);
console.log(closePosts);