Javascript

Question

我正在努力根据另一个 JSON 的值过滤一个地理 json。具体来说，我正在尝试根据另一个 json 中的 id 过滤 geojson。每次，我过滤的 geojson returns 为空。

我创建了一个 JSfiddle here 来展示我正在尝试做的事情。

密码是

    mygeojson={crs: {properties: {name: "EPSG:4326"}, type: "name"},
features: [
    {geometry: {coordinates: [25, 37], type: "Point"},properties: {"wineryid":3},type:"Feature"},
  {geometry: {coordinates: [26, 37], type: "Point"},properties: {"wineryid":45},type:"Feature"},
  {geometry: {coordinates: [25, 38], type: "Point"},properties: {"wineryid":34},type:"Feature"},
  {geometry: {coordinates: [28, 37], type: "Point"},properties: {"wineryid":42},type:"Feature"},
  {geometry: {coordinates: [25, 342], type: "Point"},properties: {"wineryid":80},type:"Feature"},
  {geometry: {coordinates: [25.24, 37.12], type: "Point"},properties: {"wineryid":120},type:"Feature"}], type:"FeatureCollection"};
mylist=[{id:1, name:"test1"},{id:34,name:"test2"}];
tempgeojson = mygeojson.features.filter(function (item) {
  return (mylist.every(f => item.properties.id === f.id) )
});
filteredgeojson={};
filteredgeojson.features=tempgeojson;
filteredgeojson.crs=mygeojson.crs;
filteredgeojson.type="FeatureCollection";

如何获得我需要的过滤后的地理位置json？（在这个例子中，预期的 geojson 是关于 id 的 1 和 34。）

Answer 1

尝试使用以下代码进行特征过滤

filteredgeojson.features = mygeojson.features.filter(item => { 
     if(mylist.filter(myitem => myitem.id === item.properties.wineryid).length > 0) { 
          return item;
      }
});

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