Google地圖V3是否有Epoly等效物？

Question

我成功安裝了用於V2的epoly.js並使其工作。

然而，在V3我得到這個錯誤在我的控制檯：

ReferenceError: Can't find variable: GPolygon

任何想法？

Answer 1

OK，我發現在http://dyve.posterous.com/?tag=googlemaps

的傢伙（S）的解決方案提供者，他們提供2個套路：

if (!google.maps.Polygon.prototype.getBounds) { 
    google.maps.Polygon.prototype.getBounds = function(latLng) { 
     var bounds = new google.maps.LatLngBounds(); 
     var paths = this.getPaths(); 
     var path; 
     for (var p = 0; p < paths.getLength(); p++) { 
      path = paths.getAt(p); 
      for (var i = 0; i < path.getLength(); i++) { 
       bounds.extend(path.getAt(i)); 
      } 
     } 
     return bounds; 
    } 
} 

if (!google.maps.Polygon.prototype.contains) { 
    google.maps.Polygon.prototype.contains = function(latLng) { 
     // Outside the bounds means outside the polygon 
     if (!this.getBounds().contains(latLng)) { 
      return false; 
     } 
     var lat = latLng.lat(); 
     var lng = latLng.lng(); 
     var paths = this.getPaths(); 
     var path, pathLength, inPath, i, j, vertex1, vertex2; 

     // Walk all the paths 
     for (var p = 0; p < paths.getLength(); p++) { 

      path = paths.getAt(p); 
      pathLength = path.getLength(); 
      j = pathLength - 1; 
      inPath = false; 
      for (i = 0; i < pathLength; i++) { 
       vertex1 = path.getAt(i); 
       vertex2 = path.getAt(j); 
       if (vertex1.lng() < lng && vertex2.lng() >= lng || vertex2.lng() < lng && vertex1.lng() >= lng) { 
        if (vertex1.lat() + (lng - vertex1.lng())/(vertex2.lng() - vertex1.lng()) * (vertex2.lat() - vertex1.lat()) < lat) { 
         inPath = !inPath; 
        } 
       } 
       j = i; 
      } 
      if (inPath) { 
       return true; 
      } 
     } 
     return false; 
    } 
} 

我再簡單地添加該代碼

function initialize() { 
    var areas = []; 
    var coords = []; 
    var latlng = new google.maps.LatLng(53.85, -2.15); 
    var myOptions = { 
    zoom: 12, 
    center: latlng, 
    mapTypeId: google.maps.MapTypeId.ROADMAP 
    }; 
    var map = new google.maps.Map(document.getElementById("map_canvas"), myOptions); 
    var marker = new google.maps.Marker({ 
     position: latlng, 
     map: map, 
     title:"Hello World!", 
     draggable: true 
    }); 
    coords[0] = [ 
     new google.maps.LatLng(53.9, -2.15), 
     new google.maps.LatLng(53.91, -2.05), 
     new google.maps.LatLng(53.85, -2.10), 
     new google.maps.LatLng(53.9, -2.15) 
    ]; 
    areas[0] = new google.maps.Polygon({ 
    paths: coords[0], 
    strokeColor: "#FF0000", 
    strokeOpacity: 0.8, 
    strokeWeight: 2, 
    fillColor: "#FF0000", 
    fillOpacity: 0.35 
    }); 
    areas[0].setMap(map); 
    coords[1] = [ 
     new google.maps.LatLng(53.9, -2.14), 
     new google.maps.LatLng(53.88, -2.05), 
     new google.maps.LatLng(53.85, -2.09), 
    ]; 
    areas[1] = new google.maps.Polygon({ 
    paths: coords[1], 
    strokeColor: "#00FF00", 
    strokeOpacity: 0.8, 
    strokeWeight: 2, 
    fillColor: "#00FF00", 
    fillOpacity: 0.35 
    }); 
    areas[1].setMap(map); 
    google.maps.event.addListener(marker, 'dragend', function(point) { 
    for(var area = 0; area <= areas.length; area++) { 
     alert(area+" "+areas[area].contains(point.latLng)); 
    } 
    }); 
} 

拖動釘住1或兩個三角形以獲得真/假。

Answer 2

是的，這裏找到一個新的移植版本： http://www.geocodezip.com/scripts/v3_epoly.js