如何根据实际情况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]);

}
}

我想在条件为时删除映射this. sensorarray .filter(sense=>sense.name == selectsensor。长度- 1])。长度比;0)

Answer 1

地图可以删除:

map.remove()

那么它看起来应该是这样的:

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