我正在尝试创建一个在地图上绘制圆圈的网络应用。如果numUsers
属性为> = 1,则圆圈为绿色,如果numUsers
属性为0,则圆圈为红色(默认值为0)。
下面是我的数据源的结构:
{
"type":"FeatureCollection",
"features":[
{
"type": "Feature",
"id": 0,
"geometry":{
"type":"Point",
"coordinates":[
1.49129,
42.46372
]
},
"properties": {
"numUsers":0
}
}
]
}
所有圆最初都渲染为红色,这是我想要的,因为每个圆的numUsers
属性最初设置为0。但是,我想通过设置将其中一个圆更改为绿色numUsers属性设置为1。我正在尝试使用setFeatureState
,但不不会将圆圈的颜色更改为绿色:
map.setFeatureState({source: "cities", id : 0}, {numUsers : 1});
下面是我的渲染JS代码:
map.on('style.load', function (e) {
map.addSource('cities', {
"type": "geojson",
"data": "cities.geojson",
"cluster": true,
"clusterMaxZoom": 14,
"clusterRadius": 80
});
map.addLayer({
"id": "cities",
"type": "circle",
"source": "cities",
"paint": {
"circle-color": {
property: 'numUsers',
stops: [
[0, '#ff6666'],
[1, '#33ff33']
]
}
}
}, 'settlement-label');
});
答案 0 :(得分:3)
您应该使用“功能状态” [1] 表达式来获取使用setFeatureState
设置的状态,并使用“ case”表达式来切换状态值并设置所需的颜色
这是要点:
// update after 2 seconds
setTimeout(() => {
map.setFeatureState({ id: 0, source: "geom" }, { numUsers: 1 });
map.setFeatureState({ id: 1, source: "geom" }, { numUsers: 2 });
}, 2000);
map.addLayer({
id: "geom",
type: "circle",
paint: {
"circle-color": [
"case",
["==", ["feature-state", "numUsers"], 1], "blue",
["==", ["feature-state", "numUsers"], 2], "green",
"red"
],
"circle-radius": 4
},
source: { /* ... source */ }
});
带有工作图的密码笔:https://codepen.io/manishraj/full/YzKeBwv
[1] https://docs.mapbox.com/mapbox-gl-js/style-spec/#expressions-feature-state
答案 1 :(得分:0)
我不确定您是否可以根据https://docs.mapbox.com/mapbox-gl-js/style-spec/#other-function使用具有功能状态的那种函数表达式。
相反,您应该可以将match
与["feature-state", "numUsers"]
一起使用<rect
inkscape:label="#rect1067"
y="50.32835"
x="154.16335"
height="25.92329"
width="23.673288"
id="lobby-link"
style="fill:none;fill-opacity:1;stroke:#000000;
stroke-width:0.32671064;stroke-miterlimit:4;
stroke-dasharray:none;stroke-opacity:1" />
表达式https://docs.mapbox.com/mapbox-gl-js/style-spec/#expressions-match来获取表达式https://docs.mapbox.com/mapbox-gl-js/style-spec/#expressions-feature-state的特征状态