raphael - 欧洲地图中的拉斐尔情节点

Question

1. 你好，

   我有一张欧洲拉斐尔地图。现在我想在地图中的某些城市上绘制点。我尝试将纬度 n 经度转换为其中的绘图点。但不幸的是它正在其他地方绘制。是不是我们应该有世界地图绘图点？这是我的代码。

   脚本类型="text/javascript" charset="utf-8">

          $(document).ready(function() {
           var rsr = Raphael('map', '631', '686');
           var attr = {
                      fill: "#C0C0C0",
                      stroke: "#666",
                      "stroke-width": 1,
                      "stroke-linejoin": "round"
                  };
              var world = {};
              world.Portugal = rsr.path("56,0.133-1.32,0.527c-0.661,1.321-0.264,2.906-  0.925,4.228c-0.528,1.057-3.698,5.415-3.434,6.868c0.132,0.526,1.056-0.529,1.584-0.529c0.792-0.132,1.585,0.133,2.377,0c0.396,0,0.792-0.396,1.188-0.264
   

   c2.113,0.527,8.981,5.019,9.906,4.887c0.396,0,4.49-1.981,4.754-2.113C57.876,621.536,58.537,621.536,59.197,621.536L59.197,621.r)r.5attr.6 .Spain = rsr.path(" M194.57,552.728c0.924,0.396,1.981,0.63.434,4.754c-,0,0.792,0 c0.661,0.133,1.453,0.133,1.849,0.528c0.66,0.528 ,0.264,1.717,0.924,2.113v0.132C190.74,552.066,190.476,553.916,194.57,552.728 L194.57,552.728z").attr(attr);

      var current = null;
               for(var country in world) {
                (function (st, country) {
                  country = country.toLowerCase();
                    st[0].style.cursor = "pointer";
                    st[0].onmouseover = function () { 
                        st.animate({fill:"#808080", stroke: "#ccc"}, 500);
                    };
                    st[0].onmouseout = function () {
                        st.animate({fill: "#C0C0C0", stroke: "#666"}, 500);
                        st.toFront();
                        R.safari();
                    };
                    st[0].onclick = function () {
                    st.toFront();
                          st.animate({
                            fill: '#808080',
                            transform: 's1.5 '
                        }, 1000);
                    };
                })(world[country], country);
              }
            });   
     var cities = {};//here i define the cities with lat n long but both draws in thesame point all time
           cities.rome = plot(55.70466,13.19101,1);
         cities.copenhagen = plot(55.676097,12.568337,1);
           var city_attr = {
                   fill:"#FF7F50",
                    stroke:"#666",
                   opacity: .3
               };
           function plot(lat,lon,size) {
               size = size * .5 + 4;
               return   rsr.circle(lon2x(lon),lat2y(lat),size).attr(city_attr);
             }
   
           function lon2x(lon) {
               var xfactor = 1.5255;
               var xoffset = 263.58;
               var x = (lon * xfactor) + xoffset;
               return x;           }           function lat2y(lat) {
                var yfactor = -1.5255;
                   var yoffset = 130.5;
                   var y = (lat * yfactor) + yoffset;
                   return y;           }
        }); 
         var myMarker = rsr.ellipse(513.859,35.333, 7, 7).attr({
             stroke: "none", 
             opacity: .7, 
             fill: "#f00"
           });
   

Answer 1

地图编码的坐标似乎相当随意。如果是这样，就没有[简单的]方法可以自动确定映射。我建议在其自己的坐标系中获取矢量图像的边界框，并在常规地图上的纬度/经度坐标中获取相应的边界框，并从中导出映射，至少作为第一近似值。