如何根据条件移除或清空传单地图

Question

我想根据所选条件删除地图。

.component.ts

dataChanged (selectsensor) {
console.log(selectsensor)
for( var j=0;j<selectsensor.length;j++){


  for (var i =0; i< this.sensors.length;i++){

    if(this.sensors[i].name == selectsensor[selectsensor.length - 1]){


      if(this.sensorsarray.filter(sense=>sense.name == selectsensor[selectsensor.length - 1]).length > 0){
      console.log(this.sensorsarray.filter(sense=>sense.name == selectsensor[selectsensor.length - 1]).length);
      for(var t=0;t<this.sensorsarray.length;t++){
        if(this.sensorsarray[t].name == selectsensor[selectsensor.length - 1]){
          this.sensorsarray.splice(t);
          
        }
      }
      console.log(this.sensorsarray);
      
        alert("Value exists!")
        
    }else{
      this.sensorsarray.push(this.sensors[i]);
      var v= "map"+ (this.sensorsarray.length - 1 )+"";
      map[i] = L.map(v).setView([this.sensors[i].latitude,this.sensors[i].longitude], 11);
        
      L.tileLayer('https://{s}.tile.openstreetmap.org/{z}/{x}/{y}.png', {
          attribution: ' <span style="font-size:17px">'+this.sensors[i].name+' </span>'
        }).addTo(map[i]);
        var latlng = L.latLng(this.sensors[i].latitude, this.sensors[i].longitude)
  
    
        marker = L.marker(latlng, { icon: sensoricon }).addTo(map[i]);
        circlemark[i] = L.circle(latlng, { radius: 5000, weight: 0 }).addTo(map[i]);


    }
    }

我想在条件为 this.sensorsarray.filter(sense=>sense.name == selectsensor[selectsensor.length - 1]).length > 0)

Answer 1

可以通过以下方式删除地图：

map.remove()

那么它应该看起来像你：

this.sensorsarray.filter(sense=>sense.name == selectsensor[selectsensor.length - 1]).length > 0){
   map[i].remove()
}

如何根据条件移除或清空传单地图

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