Question

我正在使用具有硬编码值的函数为传单地图上的多边形指定颜色。样式函数从$.getJSON内部调用。我想重写我的代码，以便直接从数据中提取值而不是硬编码 - 理论上，这样我将来可以更轻松地重用代码

$.getJSON调用的（截断的）geojson数据是：

{"type": "FeatureCollection","crs": { "type": "name", "properties": { "name": "urn:ogc:def:crs:OGC:1.3:CRS84" } },
"features": [{"type": "Feature", "properties": {"MAJORCOLOR": "ZONE A"}, "geometry": { "type": "Polygon", "coordinates": [ [ [ -77.946264290220981, 42.574112051418197 ], [ -77.954525714402251, 42.569801987122105 ], [ -77.964847297117899, 42.562124252064194 ]]]}},  
"features": [{"type": "Feature", "properties": {"MAJORCOLOR": "ZONE B"}, "geometry": { "type": "Polygon", "coordinates": [ [ [ -77.946264290220981, 42.574112051418197 ], [ -77.954525714402251, 42.569801987122105 ], [ -77.964847297117899, 42.562124252064194 ]]]}},
"features": [{"type": "Feature", "properties": {"MAJORCOLOR": "ZONE A"}, "geometry": { "type": "Polygon", "coordinates": [ [ [ -77.946264290220981, 42.574112051418197 ], [ -77.954525714402251, 42.569801987122105 ], [ -77.964847297117899, 42.562124252064194 ]]]]}}]}

$.getJSON是：

$.getJSON('data/ecozone_wgs84_multipart.geojson', function(data){
var geojsonLayer = L.geoJson(data.features, {  
    onEachFeature: makeMarkers,
    //this provides thematic styling to the layers
    style: style
})
.addTo(map);  

//call the function to create keys
getArray(data);
});

这是根据数据值

分配颜色的功能

//get color depending on the Zone value
function getColor(z) {
return  z == 'ZONE A' ? '#a6cee3':
        z == 'ZONE B' ? '#1f78b4':
        z == 'ZONE C' ? '#b2df8a':
'#000000';}

style函数从$.getJSON

内部调用

function style(feature) {
return {
    fillColor: getColor(feature.properties.MAJORCOLOR),
    color: getColor(feature.properties.MAJORCOLOR),
    weight: 1.25,
    opacity: 0.95,
    fillOpacity: 0.5
};}

我想重写getColor函数，以便不是使用硬编码的值来确定颜色，而是从我从geojson创建的数组中提取值，我已经编写如下：

//this is a function that pulls the values from MAJORCOLOR to create the array
function getArray(data) {
for (var i=0; i< data.features.length; i++) {
keys.push(data.features[i].properties.MAJORCOLOR);
}}

//this is a function to collapse to unique values
function unique(keys) {
return keys.reduce(function(p, c) {
    if (p.indexOf(c) < 0) p.push(c);
    return p;
}, []);};

然而，当我重写颜色函数以使用数组中的值而不是它不起作用时 - 多边形都被赋予“else”颜色＃000000而不是我想要的颜色。

function getColor(z) {
return  z == unique(keys)[0] ? '#a6cee3':
        z == unique(keys)[1] ? '#1f78b4':
        z == unique(keys)[2] ? '#b2df8a':
        '#000000';}

为什么这不起作用？

从控制台查看unique(keys)，我得到["ZONE A", "ZONE B", "ZONE C"]所以我知道我正在正确创建keys ...我很困惑。

提前感谢您解决我的问题！

Answer 1

var keys = [];

function getArray(data) {
  for (var i = 0; i < data.features.length; i++) {
    keys.push(data.features[i].properties.MAJORCOLOR);
  }
}

function unique(keys) {
  return keys.reduce(function(p, c) {
    if (p.indexOf(c) < 0) p.push(c);
    return p;
  }, []);
};

function getColor(z) {
  return z == unique(keys)[0] ? '#a6cee3':
         z == unique(keys)[1] ? '#1f78b4':
         z == unique(keys)[2] ? '#b2df8a': '#000000';
} 

$.getJSON('data/ecozone_wgs84_multipart.geojson', function(data){
  getArray(data);

  var geojsonLayer = L.geoJson(data.features, {  
    onEachFeature: makeMarkers,
    style: function (feature) {


      // keys avaliable
      // getColor('xxxxxx'); keys availible
    }
  }).addTo(map);  
});

函数仅适用于硬编码值

1 个答案: