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解决方案：使用相同的源创建两个不同的图层或使用 Mike 的解决方案。

有没有办法根据显示的地图设置特征的样式？我找不到在样式函数中获取地图属性的方法：

var styleBuilding = function(feature){
  console.log(feature);
  return [
  new ol.style.Style({
    stroke: new ol.style.Stroke({
      color: 'blue',
      width: 3
    }),
    fill: new ol.style.Fill({
      color: 'rgba(0, 0, 255, 0.1)'
    })
  }),
  new ol.style.Style({
    image: new ol.style.Circle({
      radius: 7,
      fill: new ol.style.Fill({
        color: 'orange'
      })
    }),
    geometry: function(feature) {
      var coordinates = feature.getGeometry().getCoordinates()[0];
      return new ol.geom.Point(coordinates[2]);
    }
  })
];
};

最好的问候蒂姆

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1 回答 1

1

视图分辨率可用作样式函数的第二个参数。任何其他视图属性（例如中心）都可以通过引用样式函数中的视图来获得，因此在此示例中，点颜色取决于该点显示在视图的哪个象限中

<!DOCTYPE html>
<html>
  <head>
    <title>Simple Map</title>
    <link rel="stylesheet" href="https://openlayers.org/en/v6.3.1/css/ol.css" type="text/css">
    <!-- The line below is only needed for old environments like Internet Explorer and Android 4.x -->
    <script src="https://cdn.polyfill.io/v2/polyfill.min.js?features=requestAnimationFrame,Element.prototype.classList,URL"></script>
    <script src="https://openlayers.org/en/v6.3.1/build/ol.js"></script>
    <style>
      html, body, .map {
        margin: 0;
        padding: 0;
        width: 100%;
        height: 100%;
      }
    </style>
  </head>
  <body>
    <div id="map" class="map"></div>
    <script>
      var map = new ol.Map({
        layers: [
          new ol.layer.Tile({
            source: new ol.source.OSM()
          }),
          new ol.layer.Vector({
            source: new ol.source.Vector({
              features: [
                new ol.Feature(new ol.geom.Point([1e6, 1e6])),
                new ol.Feature(new ol.geom.Point([1e6, -1e6])),
                new ol.Feature(new ol.geom.Point([-1e6, -1e6])),
                new ol.Feature(new ol.geom.Point([-1e6, 1e6])),
              ]
            }),
            style: function(feature){
              let color;
              const coordinate = feature.getGeometry().getCoordinates();
              const viewCenter = map.getView().getCenter();
              if (coordinate[0] < viewCenter[0]){
                if (coordinate[1] < viewCenter[1]){
                  color = 'red';
                } else {
                  color = 'orange';
                }
              } else {
                if (coordinate[1] < viewCenter[1]){
                  color = 'green';
                } else {
                  color = 'blue';
                }
              }
              return [
                new ol.style.Style({
                  image: new ol.style.Circle({
                    radius: 7,
                    fill: new ol.style.Fill({
                      color: color
                    })
                  })
                })
              ];
            }
          })
        ],
        target: 'map',
        view: new ol.View({
          center: [0, 0],
          zoom: 2
        })
      });
    </script>
  </body>
</html>

如果地图共享一个源并查看一个图层，则需要为每个地图定义一个图层，因此您可以创建一个共享样式函数，该函数接收地图（或其他一些 ID）以指示正在为哪个地图设置样式

  var map1 = new ol.Map({
    layers: [
      new ol.layer.Vector({
        source: sharedSource,
        style: function(feature){
          return sharedStyleFunction(feature, map1)
        }
      })
    ],
    target: 'map1',
    view: sharedView
  });

  var map2 = new ol.Map({
    layers: [
      new ol.layer.Vector({
        source: sharedSource,
        style: function(feature){
          return sharedStyleFunction(feature, map2)
        }
      })
    ],
    target: 'map2',
    view: sharedView
  });
于 2020-06-23T14:58:40.900 回答