Google地图多重叠加无累积不透明度

Question

我有一张地图上有多个圆圈相互交叉(下面是一个只有两个圆圈的例子,但至少有100个圆圈).当它们交叉时,不透明度加倍,所以当我有一个5或6个圆圈之间的交叉时,它只会变成大约100%的不透明度.

有没有办法让第二个圆圈不显示"第一个圆圈"？实际上一个人不这么认为,但也许有人已经预料到这样的事......

左:我有什么--------------------------------------------- - 对:我想要的

以防万一你想玩:http: //jsfiddle.net/ZWt6w/

var populationOptions = {
      strokeWeight: 0,
      fillColor: '#FF0000',
      fillOpacity: 0.5,
      map: map,
      center: citymap[city].center,
      radius: citymap[city].population
    };
    // Add the circle for this city to the map.
    cityCircle = new google.maps.Circle(populationOptions);

谢谢你的帮助 ;)

Answer 1

使用Polygon类可以使重叠不透明度变为单一.

var peanut = new google.maps.Polygon({
                 paths: [drawCircle(citymap['chicago'].center, citymap['chicago'].population/3000, 1),//division by 3000 to suit
                        drawCircle(citymap['losangeles'].center,citymap['losangeles'].population/3000, 1)],
                 strokeColor: "#ff0000",
                 strokeOpacity: 0.35,
                 strokeWeight: 0,
                 fillColor: "#FF0000",
                 fillOpacity: 0.35
     });
     peanut.setMap(map);

Answer 2

使用多个路径绘制一个多边形----------绘制多个圆圈

@david strachan的答案解决了我的问题的很大一部分.以下是此解决方案的一部分:首先,您必须使用此"drawCircle"函数,而不是Google Maps API V3的Circle对象:

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
    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
    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)); 
        ey = point.lng() + (rlng * Math.cos(theta)); // center a + radius x * cos(theta) 
        ex = point.lat() + (rlat * Math.sin(theta)); // center b + radius y * sin(theta) 
        extp.push(new google.maps.LatLng(ex, ey));
    }
    return extp;
}

此函数返回路径,因此您可以使用它来建立一个路径数组,您将在构建单个Polygon对象后使用它们:

var polys = [] ;
$(xml).find("trkpt").each(function() { // Parsing every points of my track
    var p = new google.maps.LatLng($(this).attr("lat"), $(this).attr("lon"));
    points.push(p);
    if ( ( i++ % 10 ) == 0 ) // Only display a circle every 10 points
    {
        polys.push(drawCircle(p,radius/1609.344,1)) ; // Radius value is in meters for me, so i divide to make it in miles
    }
});


peanutcircle = new google.maps.Polygon({
    paths: polys,
    strokeOpacity: 0,
    strokeWeight: 0,
    fillColor: color,
    fillOpacity: 0.35,
});
peanutcircle.setMap(map);

这就是全部,你绘制了一个复杂但单一的多边形,可能更容易使用.

对我来说唯一的问题是检查这个单一多边形中包含的标记(使用google function containsLocation和github.com/tparkin/Google-Maps-Point-in-Polygon)效果不佳,所以我不得不继续使用我的倍数圆来检查标记是否在我的区域中.

感谢@david strachan的回答.