更改Google Maps JavaScript API v3圈外的地图不透明度

时间:2013-11-01 22:28:28

标签: javascript google-maps google-maps-api-3

所以我目前在地图上放了一个圆圈:

    var optionsCercle = {
        center: latlang,
        map: map,
        radius: 1000,
        fillOpacity: 0.1,
        strokeWeight: 0
    };
    this.circ = new google.maps.Circle(optionsCercle);

现在我将圆圈的内部填充到容量:0.1,但我想要做的是将所有内容放在地图上,除了圆圈内部以填充容量:0.1。我想要在视口中的所有内容,除了圆圈“模糊”。我怎样才能做到这一点?

3 个答案:

答案 0 :(得分:20)

在纽约市上放置一个圆形“洞”的例子(由this Google Example修改):

var citymap = {};
citymap['newyork'] = {
  center: new google.maps.LatLng(40.714352, -74.005973),
  population: 8143197
};

var cityCircle;
var bounds = new google.maps.LatLngBounds();

function drawCircle(point, radius, dir) { 
  var d2r = Math.PI / 180;   // degrees to radians 
  var r2d = 180 / Math.PI;   // radians to degrees 
  var earthsradius = 3963; // 3963 is the radius of the earth in miles use 6371 if using kilometers
  var points = 32; 

  // find the raidus in lat/lon 
  var rlat = (radius / earthsradius) * r2d; 
  var rlng = rlat / Math.cos(point.lat() * d2r); 

  var extp = new Array(); 
  if (dir==1)   {var start=0;var end=points+1} // one extra here makes sure we connect the ends
  else      {var start=points+1;var end=0}
  for (var i=start; (dir==1 ? i < end : i > end); i=i+dir) { 
    var theta = Math.PI * (i / (points/2)); 
    var ey = point.lng() + (rlng * Math.cos(theta)); // center a + radius x * cos(theta) 
    var ex = point.lat() + (rlat * Math.sin(theta)); // center b + radius y * sin(theta) 
    extp.push(new google.maps.LatLng(ex, ey)); 
    bounds.extend(extp[extp.length-1]);
  } 
  return extp;
}

function initialize() {
  // Create the map.
  var mapOptions = {
    zoom: 4,
    center: new google.maps.LatLng(37.09024, -95.712891),
    mapTypeId: google.maps.MapTypeId.TERRAIN
  };

  var map = new google.maps.Map(document.getElementById('map-canvas'),
      mapOptions);

  var outerbounds = [ // covers the (mercator projection) world
        new google.maps.LatLng(85,180),
    new google.maps.LatLng(85,90),
    new google.maps.LatLng(85,0),
    new google.maps.LatLng(85,-90),
    new google.maps.LatLng(85,-180),
    new google.maps.LatLng(0,-180),
    new google.maps.LatLng(-85,-180),
    new google.maps.LatLng(-85,-90),
    new google.maps.LatLng(-85,0),
    new google.maps.LatLng(-85,90),
    new google.maps.LatLng(-85,180),
    new google.maps.LatLng(0,180),
    new google.maps.LatLng(85,180)];

    // options for the polygon
    var populationOptions = {
      strokeColor: '#FF0000',
      strokeOpacity: 0.8,
      strokeWeight: 2,
      fillColor: '#FF0000',
      fillOpacity: 0.35,
      map: map,
      paths: [outerbounds,drawCircle(citymap['newyork'].center,10,-1)]
    };
    // Add the circle for this city to the map.
    cityCircle = new google.maps.Polygon(populationOptions);
    map.fitBounds(bounds);
}

google.maps.event.addDomListener(window, 'load', initialize);

Working example

screen shot

答案 1 :(得分:0)

{-Lat, (Long-180)%180, 20037508.34-Radius}对我来说很好,Thx到Bahram Ardalan。这是我的代码:

lat = 53;
lng = 10;
distance = 10 * 1000; // 10km

new google.maps.Circle({
                strokeColor: '#FF0000',
                strokeOpacity: 0.8,
                strokeWeight: 2,
                fillColor: '#FF0000',
                fillOpacity: 0.35,
                map: this.map,
                center: {lat: -1 * lat, lng: (lng - 180 ) % 180},
                radius: 20037508.34 - distance      

答案 2 :(得分:-1)

在相反的位置绘制一个圆,像

一样绘制负半径
{-Lat, (Long-180)%180, 20037508.34-Radius}