OpenLayers 3矢量图层上所有要素的范围？

Question

我有一个OpenLayers 3地图,带有基础层和矢量图层.

this.topoLayer = new ol.layer.Vector({
  source: new ol.source.Vector(),
  style: style
});

var baseLayer = new ol.layer.Tile({
    source: new ol.source.XYZ({
        url: 'http://[…]/{z}/{x}/{y}.png',
        crossOrigin: 'null'
    })
});

this.map = new ol.Map({
    target: 'map',
    layers: [baseLayer, this.topoLayer],
    view: new ol.View2D({
        center: ol.proj.transform([11.38,  48.54], this.options.markerEPSG, this.options.mapEPSG),
        zoom: 5,
    }),
});

在用户交互时,我向向量层添加和删除了几个要素.以下是添加新功能的功能:

  var feature = new ol.Feature({
        topo: topo,
        selected: false,
        geometry: new ol.geom.Point(ol.proj.transform(location, this.options.markerEPSG, this.options.mapEPSG)),
  });

  this.topoLayer.getSource().addFeatures([feature]);

添加/删除新功能后,我想自动缩放和平移地图以适应我的功能.在旧的OpenLayers API中getDataExtent,矢量图层上有一个函数可以检索显示的所有特征周围的"边界框".但我想知道如何使用新的API做到这一点.

Answer 1

在3.7版本,ol.View.fitExtent()并且ol.View.fitGeometry()已经统一在一个单一的功能:fit.

所以现在的代码是:

 var extent = myLayer.getSource().getExtent();
 map.getView().fit(extent, map.getSize());

Answer 2

根据Tyler DeWitt的回答,您应该能够做到这样的事情:

var view = yourmap.getView();
var view2D = view.getView2D();
var extent = yourlayer.getSource().getExtent();

view2D.fitExtent(extent, yourmap.getSize());

编辑:就像蒂姆和斯图尔特说的那样,以上不再适用了.以下应该做的伎俩:

 var extent = yourlayer.getSource().getExtent();
 map.getView().fitExtent(extent, map.getSize());

Answer 3

var source = layer.getSource();
goog.events.listen(source, ol.source.VectorEventType.ADD, function(evt) {
     // taken from ol/source/vectorsource.js
     var extent = ol.extent.createEmpty();
     var geometry = null;
     var feature;
     var features = source.getFeatures().getArray();
     for (var i = 0, ii = features.length; i < ii; ++i) {
         feature = features[i];
         geometry = feature.getGeometry();
         if (!goog.isNull(geometry)) {
               ol.extent.extend(extent, geometry.getExtent());
     }
     // here you should have your extent
    }
}, false, this);

未经测试。

应该有更优雅的方式，但我现在不确定在哪里可以找到它。