Javascript 在传单热图中的层之间交换
我想在传单中的层之间动态交换。所以我认为隐藏层。这是我的密码Javascript 在传单热图中的层之间交换,javascript,leaflet,Javascript,Leaflet,我想在传单中的层之间动态交换。所以我认为隐藏层。这是我的密码 <div id="map"></div> <div> <button onclick='fun1()'>yellowhide</button> <button onclick='fun2()'>redhide</button> </div> <scrip
<div id="map"></div>
<div>
<button onclick='fun1()'>yellowhide</button>
<button onclick='fun2()'>redhide</button>
</div>
<script>
//same lat long means overlapping
addressPoints1= [ [-37.8210922667, 175.2209316333, "2"], [-37.8210819833, 175.2213903167, "3"], [-37.8210881833, 175.2215004833, "3"], [-37.8211946833, 175.2213655333, "1"]] addressPoints2= [ [-37.8210922667, 175.2209316333, "2"], [-37.8210819833, 175.2213903167, "3"], [-37.8210881833, 175.2215004833, "3"], [-37.8211946833, 175.2213655333, "1"]]
var tiles=L.tileLayer('http://{s}.tiles.mapbox.com/v3/{id}/{z}/{x}/{y}.png', { attribution: '', id: 'examples.map-20v6611k' }) var map = L.map('map', { center: [-37.87, 175.475], zoom: 16, layers: [tiles] }); var heat1 = L.heatLayer(addressPoints1,{ gradient: {1:"red"} }).addTo(map); var heat2 = L.heatLayer(addressPoints2,{ gradient: {1:"yellow"} }).addTo(map);
function fun1(){ console.log("hide yellow layer");
$('.heat2').hide(); } function fun2(){ console.log("hide red layer");
$('.heat1').hide();
}
黄皮
红皮
//相同的lat-long表示重叠
地址点1=[-37.8210922667,175.2209316333,“2”],[-37.8210819833,175.2213903167,“3”],[-37.8210881833,175.2215004833,“3”],[-37.8211946833,175.2213655333,“1”]]地址点2=[-37.8210922667,175.2209316333,“2”],[-37.8210819833,175.2213903167,“3”],[-37.82108813,175.221350833”,[-37.8211946833, 175.2213655333, "1"]]
var tiles=L.tileLayer('http://{s}.tiles.mapbox.com/v3/{id}/{z}/{x}/{y}.png',{attribute:'',id:'examples.map-20v6611k'})var map=L.map('map',{center:[-37.87175.475],zoom:16,layers:[tiles]});var heat1=L.heatLayer(addressPoints1,{gradient:{1:{red red}).addTo(map);var heat2=L.heatLayer(地址点2,{梯度:{1:“黄色”}}).addTo(地图);
函数fun1(){console.log(“隐藏黄色层”);
$('.heat2').hide();}函数fun2(){console.log(“隐藏红色层”);
$('.heat1').hide();
}
但是它不起作用。没有必要对这些图层使用jQuery和query,因为您以前已经声明为变量
heat1
和heat2
。您可以使用这些变量将它们从地图中删除或再次添加。此时,您正在使用图层的addTo
方法将图层添加到地图中:
var heat1 = L.heatLayer().addTo(map);
它在后台执行映射实例的addLayer
方法:
map.addLayer(heat1);
地图实例还具有用于从地图中删除图层的removeLayer
方法:
map.removeLayer(heat1);
因此,您可以在onclick处理程序中使用它:
function fun1 () {
map.removeLayer(heat1);
}
这里是L.Map所有图层方法的参考:如果您还没有找到解决方案,请查看 使用这些控件,可以更改地图的可见图层
var baseLayers = {
"Mapbox": tiles
};
var overlays = {
"Fancy Heatmap": heat1,
"Another fancy Heatmap": heat2
};
L.control.layers(baseLayers, overlays).addTo(map);
HTH,Meykel我不想添加或删除图层。只需在图层之间交换..就像我们有两个按钮1用于红色图层,2用于黄色图层…当我单击红色按钮时,红色图层显示在黄色图层的顶部。当我单击黄色按钮时,黄色图层显示在红色图层的顶部..在你的问题中,你说“隐藏”,意思是“放置或隐藏在视线之外;隐藏在视图之外”。你应该在你的问题中提到这一点。@iH8是的,你是对的(某种程度上)…但是hide()和remove()之间的区别是:--hide()将匹配元素的CSS显示属性设置为none。remove()从DOM中完全删除匹配的元素。