如何从JSON数据中使用经度和纬度查找最近的位置 [英] How to find nearest location using latitude and longitude from a json data

问题描述

我正在尝试创建一个网站，询问用户的位置，然后使用GeoLocation从该位置查找最近的位置(半径100m半径)，并以HTML显示结果.

I'm trying to make a website that ask for user's location then find the closest location (100m radius) from it's position using GeoLocation and display the result in HTML.

我尝试过的.

$.getJSON("places.json", function (data) {
        for (var i = 0; i < data.length; i++) {
            if ((data[i].lat - poslat) > 0.00200 || (data[i].lng - poslng) > 0.00200) {
            return data[i];
        }

        html += '<p>' + data[i].location + ' - ' + data[i].code + '</p>';
        $('#nearbystops').append(html);
    }
});

places.json

places.json

[
{
"code": "0001",
"lat": "1.28210155945393",
"lng": "103.81722480263163",
"location": "Stop 1"
},
{
"code": "0003",
"lat": "1.2777380589964",
"lng": "103.83749709165197",
"location": "Stop 2"
},
{
"code": "0002",
"lat": "1.27832046633393",
"lng": "103.83762574759974",
"location": "Stop 3"
}
]

先谢谢您！ :)

推荐答案

要计算两个坐标之间的距离，不能只减去这些值.很好，但是可以为您提供正方形内的坐标.这可能是合适的，但是大多数人确实倾向于按半径搜索位置.此功能可以做到这一点...

To calculate distance between two co-ordinates, you can't just subtract the values. That's fine but it gives you the co-ordinates that are within a square. This may be suitable but mostly people do tend to want to search locations by radius. This function will do that...

function distance(lat1, lon1, lat2, lon2, unit) {
    var radlat1 = Math.PI * lat1/180
    var radlat2 = Math.PI * lat2/180
    var theta = lon1-lon2
    var radtheta = Math.PI * theta/180
    var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
    if (dist > 1) {
        dist = 1;
    }
    dist = Math.acos(dist)
    dist = dist * 180/Math.PI
    dist = dist * 60 * 1.1515
    if (unit=="K") { dist = dist * 1.609344 }
    if (unit=="N") { dist = dist * 0.8684 }
    return dist
}

这是我从这里复制的普通代码...

It's a common piece of code which I copied from here...

https://www.geodatasource.com/developers/javascript

这里是您的示例中使用的...

And here it is, used in your example...

function distance(lat1, lon1, lat2, lon2, unit) {
    	var radlat1 = Math.PI * lat1/180
    	var radlat2 = Math.PI * lat2/180
    	var theta = lon1-lon2
    	var radtheta = Math.PI * theta/180
    	var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
    	if (dist > 1) {
    		dist = 1;
    	}
    	dist = Math.acos(dist)
    	dist = dist * 180/Math.PI
    	dist = dist * 60 * 1.1515
    	if (unit=="K") { dist = dist * 1.609344 }
    	if (unit=="N") { dist = dist * 0.8684 }
    	return dist
    }

    var data = [{
        "code": "0001",
        "lat": "1.28210155945393",
        "lng": "103.81722480263163",
        "location": "Stop 1"
    }, {
        "code": "0003",
        "lat": "1.2777380589964",
        "lng": "103.83749709165197",
        "location": "Stop 2"
    }, {
        "code": "0002",
        "lat": "1.27832046633393",
        "lng": "103.83762574759974",
        "location": "Stop 3"
    }];

    var html = "";
    var poslat = 1.28210155945393;
    var poslng = 103.81722480263163;

    for (var i = 0; i < data.length; i++) {
        // if this location is within 0.1KM of the user, add it to the list
        if (distance(poslat, poslng, data[i].lat, data[i].lng, "K") <= 0.1) {
            html += '<p>' + data[i].location + ' - ' + data[i].code + '</p>';
        }
    }

    $('#nearbystops').append(html);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="nearbystops"></div>

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