直指點:我正在構建一個帶有通過JSON拉取的自定義標記的gmap。我還需要提供搜索功能;與輸入的郵編(AU)最近的標記。地圖API v3地理編碼
標記顯示(和陣列中的其他數據)工作正常,這是我的屁股踢了第二部分。我對JS不太瞭解。目前我所要做的只是返回輸入的郵編的緯度和長度,但如果任何人都可以告訴我如何執行整個功能,那很好。任何人都可以將我指向正確的方向嗎? JS和下面的標記,並提前感謝您的任何幫助。
的JavaScript:
function initialize() {
/* Map setup */
var latlng = new google.maps.LatLng(XXX, XXX);
var mapOptions = {
center: latlng,
zoom: 16,
mapTypeId: google.maps.MapTypeId.ROADMAP
};
/* Map initialise */
var map = new google.maps.Map(document.getElementById("map-canvas"),
mapOptions);
/* set iconpath */
var iconBase = '/path-to-icons';
/* create a separate shadow icon */
var markerShadow = {
url: iconBase + 'icon_map-shadow.png',
anchor: new google.maps.Point(15, 25)
};
/* Get JSON */
jQuery.getJSON('/path-to/markers.json', function (mapdata) {
$.each(mapdata, function (key, data) {
var latlng = new google.maps.LatLng(data.lat, data.lng);
var marker = new google.maps.Marker({
position: latlng,
map: map,
icon: iconBase + 'icon_map.png',
shadow: markerShadow,
address: data.address,
subaddress: data.subaddress,
title: data.title
});
/* For the showing of the details */
// TODO : MAKE THIS NEATER
google.maps.event.addListener(marker, 'click', function() {
$("#map-details .inner").hide();
$("#map-details .inner span").html("");
$("#map-details .inner .neareststore").html(marker.title);
$("#map-details .inner .address").html(marker.address);
$("#map-details .inner .sub-address").html(marker.subaddress);
$("#map-details .inner").fadeIn(200);
});
});
});
};
google.maps.event.addDomListener(window, 'load', initialize);
function geocode() {
$("#search").submit(function(e){
e.preventDefault();
var currentloc = $("#postcode").val();
var url = "http://maps.googleapis.com/maps/api/geocode/json?address="+currentloc+"+AU&sensor=false";
$.getJSON(url, function(data) {
//JSON function
});
};
};
基本的HTML表單:
<form action="" method="post" id="search">
<fieldset>
<input name="postcode" id="postcode">
<input type="submit" id="postcodesearch" value="search">
</fieldset>
</form>
*編輯* - 已回答 不能回答我的問題呢,但有使用這裏鏈接的Haversine公式解決了這個問題:Google Maps Api v3 - find nearest markers
它需要一些整理,但這是爲了後代的要點。感謝所有的幫助。
function rad(x) {return x*Math.PI/180;}
function find_closest_marker(_lat , _lng ) {
var lat = _lat;
var lng = _lng;
var R = 6371; // radius of earth in km
var distances = [];
var closest = -1;
for(i=0;i<map.markers.length; i++) {
var mlat = map.markers[i].position.lat();
var mlng = map.markers[i].position.lng();
var dLat = rad(mlat - lat);
var dLong = rad(mlng - lng);
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(rad(lat)) * Math.cos(rad(lat)) * Math.sin(dLong/2) * Math.sin(dLong/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
distances[i] = d;
if (closest == -1 || d < distances[closest]) {
closest = i;
}
}
var closest = map.markers[closest];
map.setCenter(closest.getPosition());
$("#map-details .inner").hide();
$("#map-details .inner span").html("");
$("#map-details .inner .neareststore").html(closest.title);
$("#map-details .inner .address").html(closest.address);
$("#map-details .inner .sub-address").html(closest.subaddress);
$("#map-details .inner").fadeIn(200);
}
$(document).ready(function(){
$("#search").submit(function(e){
e.preventDefault();
var currentloc = $("#postcode").val();
var url = "http://maps.googleapis.com/maps/api/geocode/json?address="+currentloc+"+AU&sensor=false";
$.getJSON(url, function(data) {
var lat = data.results[0].geometry.location.lat;
var lng = data.results[0].geometry.location.lng;
find_closest_marker(lat,lng)
});
});
});
您是否獲得某種試圖返回輸入的緯度和經度時的錯誤? – summea
我收到了「意外標記}」,這意味着我在某處有語法錯誤,但找不到它。 –
查看[google.maps.geometry.spherical.computeDistanceBetween方法](https://developers.google.com/maps/documentation/javascript/reference#spherical)(而不是自己編寫代碼) – geocodezip