OpenLayers 3根据属性选择并高亮feature

OpenLayers 3 版本:OpenLayers v4.3.2

想通过属性查询feature,但是在ol3中并没有发现类似ol2中的getFeatureByAttribute()....

自己写函数实现,实现方法如下:

var layer = new ol.layer.Vector({
        source: new ol.source.Vector({
            features:ol.Feature,
            url: './resources/data/test.geojson',
            format: new ol.format.GeoJSON()
        }),
        style:polygonStyle
    });
    map.addLayer(layer);
function selectByAttribute(){
    var features = layer.getSource().getFeatures();
    //console.log(features);
    var value = '115街坊';
    var property = 'name';
var selectedByAttriFeature;//实际应用中设置成全局变量
for (var i = 0, ii = features.length; i < ii; i++) { if (features[i].get(property) === value) { selectedByAttriFeature = features[i]; break; } } selectedByAttriFeature.setStyle(featureSelectStyle);//高亮查询到的feature //console.log(selectedByAttriFeature.getGeometry().A);//后台输出发现,geometry属性中有一个属性A其实是中心点坐标,但是没有提供获取中心点的方法,.getGeometry().A也可获得中心点 //console.log(getCenterOfExtent(selectedByAttriFeature.getGeometry().getExtent()));//结果同上 var my_view = new ol.View({ center: getCenterOfExtent(selectedByAttriFeature.getGeometry().getExtent()), zoom: 14 }); map.setView(my_view); }

获得feature的中心点坐标:【偶然发现ol3有方法获得中心点坐标:ol.extent.getCenter(selectedFeature.getGeometry().getExtent())

function getCenterOfExtent(Extent){
    var X = Extent[0] + (Extent[2]-Extent[0])/2;
    var Y = Extent[1] + (Extent[3]-Extent[1])/2;
    return [X, Y];
}

清除已经选择的高亮feature:

function clearSelectByAttribute(){
    if(selectedByAttriFeature != null){
        selectedByAttriFeature.setStyle(null);
        selectedByAttriFeature = null;
    }
}

 

posted @ 2017-09-05 11:58  marost  阅读(14504)  评论(0编辑  收藏  举报