如何在其上点击时，我得到了一份传单形状的层类型？(How do I get a Leaflet sh

My Leaflet project allows the users to draw shapes (lines, rectangles and polygons). The user can click on the shapes to get their statistics (area, perimeter, etc).

I tried a click event on my FeatureGroup(), where all the shapes I have drawn is added. I am not sure if this is the best approach. Then upon click, a function taking an event is called. The layer type is inferred from the event object.


 //Handlers for when drawn shapes are clicked
        editableLayers.on('click', onLayerClick);

        function onLayerClick(e)
        {
            let type = e.layerType,
                layer = e.layer;
            if (type === 'polygon') {
                polygons.push(e.layer);
                let area = L.GeometryUtil.geodesicArea(layer.getLatLngs()[0]);
                console.log("New polygon area: " + area);
            }

            if (type === 'rectangle') {
                rectangles.push(e.layer);
                let area = L.GeometryUtil.geodesicArea(layer.getLatLngs()[0]);
                console.log("New rectangle area: " + area);
            }     
        }

The type object returns an undefined, and the layer object returns a bunch of parameters making no reference to the shape type. Because of that, I cannot infer the shape type and perform the correct calculations to get their stats.

Answer 1:

我会充分利用instanceof运算符此任务，例如：

function onLayerClick(ev) {
  var targetLayer = ev.sourceTarget;
  if (targetLayer instanceof L.Rectangle) {
     // Do something
  } else if (targetLayer instanceof L.Polygon) {
     // Do something
  } else if (targetLayer instanceof L.Polyline) {
     // Do something
  } else {
     // Do something
  }
}

需要注意的是，由于继承是如何工作的，任何情况下L.Rectangle也是一个实例L.Polygon ，并L.Polyline和L.Path -这就是为什么代码应该首先检查最具体的子类。

记住要检查出的单张参考，专门为位，告诉你关于继承树，例如：