43

从地理编码器结果中获取不同数组内容时遇到问题。

item.formatted_address 有效，但 item.address_components.locality 无效？

geocoder.geocode( {'address': request.term }, function(results, status) {

        response($.map(results, function(item) {

        alert(item.formatted_address+" "+item.address_components.locality)
    }            
});

// 返回的数组是;

 "results" : [
      {
         "address_components" : [
            {
               "long_name" : "London",
               "short_name" : "London",
               "types" : [ "locality", "political" ]
            } ],
          "formatted_address" : "Westminster, London, UK" // rest of array...

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4

14 回答 14

59

最终使用：

var arrAddress = item.address_components;
var itemRoute='';
var itemLocality='';
var itemCountry='';
var itemPc='';
var itemSnumber='';

// iterate through address_component array
$.each(arrAddress, function (i, address_component) {
    console.log('address_component:'+i);

    if (address_component.types[0] == "route"){
        console.log(i+": route:"+address_component.long_name);
        itemRoute = address_component.long_name;
    }

    if (address_component.types[0] == "locality"){
        console.log("town:"+address_component.long_name);
        itemLocality = address_component.long_name;
    }

    if (address_component.types[0] == "country"){ 
        console.log("country:"+address_component.long_name); 
        itemCountry = address_component.long_name;
    }

    if (address_component.types[0] == "postal_code_prefix"){ 
        console.log("pc:"+address_component.long_name);  
        itemPc = address_component.long_name;
    }

    if (address_component.types[0] == "street_number"){ 
        console.log("street_number:"+address_component.long_name);  
        itemSnumber = address_component.long_name;
    }
    //return false; // break the loop   
});
于 2011-06-21T14:44:42.257 回答
15

尝试了几个不同的请求：

MK107BX

英国克利夫兰公园新月

就像你说的，返回的数组大小不一致，但两个结果的 Town 似乎都在类型为 [“locality”，“political”] 的 address_component 项中。也许您可以将其用作指标？

编辑：使用 jQuery 获取 locality 对象，将其添加到您的响应函数中：

var arrAddress = item.results[0].address_components;
// iterate through address_component array
$.each(arrAddress, function (i, address_component) {
    if (address_component.types[0] == "locality") // locality type
        console.log(address_component.long_name); // here's your town name
        return false; // break the loop
    });
于 2011-06-15T15:22:49.417 回答
10

当用户单击地图上的某个位置时，我必须创建一个程序来填写用户表单中的纬度、经度、城市、县和州字段。该页面可以在http://krcproject.groups.et.byu.net找到，它是一个用户表单，允许公众为数据库做出贡献。我不声称自己是专家，但它工作得很好。

<script type="text/javascript">
  function initialize() 
  {
    //set initial settings for the map here
    var mapOptions = 
    {
      //set center of map as center for the contiguous US
      center: new google.maps.LatLng(39.828, -98.5795),
      zoom: 4,
      mapTypeId: google.maps.MapTypeId.HYBRID
    };

    //load the map
    var map = new google.maps.Map(document.getElementById("map"), mapOptions);

    //This runs when the user clicks on the map
    google.maps.event.addListener(map, 'click', function(event)
    {
      //initialize geocoder
      var geocoder = new google.maps.Geocoder()

      //load coordinates into the user form
      main_form.latitude.value = event.latLng.lat();
      main_form.longitude.value = event.latLng.lng();

      //prepare latitude and longitude
      var latlng = new google.maps.LatLng(event.latLng.lat(), event.latLng.lng());

      //get address info such as city and state from lat and long
      geocoder.geocode({'latLng': latlng}, function(results, status) 
      {
        if (status == google.maps.GeocoderStatus.OK) 
        {
          //break down the three dimensional array into simpler arrays
          for (i = 0 ; i < results.length ; ++i)
          {
            var super_var1 = results[i].address_components;
            for (j = 0 ; j < super_var1.length ; ++j)
            {
              var super_var2 = super_var1[j].types;
              for (k = 0 ; k < super_var2.length ; ++k)
              {
                //find city
                if (super_var2[k] == "locality")
                {
                  //put the city name in the form
                  main_form.city.value = super_var1[j].long_name;
                }
                //find county
                if (super_var2[k] == "administrative_area_level_2")
                {
                  //put the county name in the form
                  main_form.county.value = super_var1[j].long_name;
                }
                //find State
                if (super_var2[k] == "administrative_area_level_1")
                {
                  //put the state abbreviation in the form
                  main_form.state.value = super_var1[j].short_name;
                }
              }
            }
          }
        }
      });
    });
  }
</script>
于 2012-12-03T08:59:37.217 回答
6

我假设您想获得城市和州/省：

var map_center = map.getCenter();
reverseGeocode(map_center);


function reverseGeocode(latlng){
  geocoder.geocode({'latLng': latlng}, function(results, status) {
      if (status == google.maps.GeocoderStatus.OK) {
            var level_1;
            var level_2;
            for (var x = 0, length_1 = results.length; x < length_1; x++){
              for (var y = 0, length_2 = results[x].address_components.length; y < length_2; y++){
                  var type = results[x].address_components[y].types[0];
                    if ( type === "administrative_area_level_1") {
                      level_1 = results[x].address_components[y].long_name;
                      if (level_2) break;
                    } else if (type === "locality"){
                      level_2 = results[x].address_components[y].long_name;
                      if (level_1) break;
                    }
                }
            }
            updateAddress(level_2, level_1);
       } 
  });
}

function updateAddress(city, prov){
   // do what you want with the address here
}

不要尝试返回结果，因为您会发现它们是未定义的——异步服务的结果。您必须调用一个函数，例如 updateAddress();

于 2013-04-08T15:13:32.830 回答
5

我创建了这个函数来获取地理编码器结果的主要信息：

const getDataFromGeoCoderResult = (geoCoderResponse) => {
  const geoCoderResponseHead = geoCoderResponse[0];
  const geoCoderData = geoCoderResponseHead.address_components;
  const isEmptyData = !geoCoderResponseHead || !geoCoderData;

  if (isEmptyData) return {};

  return geoCoderData.reduce((acc, { types, long_name: value }) => {
    const type = types[0];

    switch (type) {
      case 'route':
        return { ...acc, route: value };
      case 'locality':
        return { ...acc, locality: value };
      case 'country':
        return { ...acc, country: value };
      case 'postal_code_prefix':
        return { ...acc, postalCodePrefix: value };
      case 'street_number':
        return { ...acc, streetNumber: value };
      default:
        return acc;
    }
  }, {});
};

因此，您可以像这样使用它：

const geoCoderResponse = await geocodeByAddress(value);
const geoCoderData = getDataFromGeoCoderResult(geoCoderResponse);

假设您要进行搜索Santiago Bernabéu Stadium，那么结果将是：

{
  country: 'Spain',
  locality: 'Madrid',
  route: 'Avenida de Concha Espina',
  streetNumber: '1',
}
于 2020-04-26T05:38:28.813 回答
4

这对我有用：

const localityObject = body.results[0].address_components.filter((obj) => {
  return obj.types.includes('locality');
})[0];
const city = localityObject.long_name;

或一口气：

const city = body.results[0].address_components.filter((obj) => {
  return obj.types.includes('locality');
)[0].long_name;

我在 Node 中这样做，所以这没关系。如果您需要支持 IE，则需要使用 polyfillArray.prototype.includes或找到另一种方法。

于 2018-07-22T14:46:42.957 回答
3

我认为谷歌没有提供某种功能来获得这些是一个真正的痛苦。无论如何，我认为找到正确对象的最佳方法是：

geocoder.geocode({'address': request.term }, function(results, status){

   response($.map(results, function(item){

      var city = $.grep(item.address_components, function(x){
         return $.inArray('locality', x.types) != -1;
      })[0].short_name;

      alert(city);
   }            
});
于 2015-01-12T08:41:58.220 回答
1 
// Use Google Geocoder to get Lat/Lon for Address
function codeAddress() {
    // Function geocodes address1 in the Edit Panel and fills in lat and lon
    address = document.getElementById("tbAddress").value;
    geocoder.geocode({ 'address': address }, function (results, status) {
        if (status == google.maps.GeocoderStatus.OK) {
            loc[0] = results[0].geometry.location.lat();
            loc[1] = results[0].geometry.location.lng();
            document.getElementById("tbLat").value = loc[0];
            document.getElementById("tbLon").value = loc[1];
            var arrAddress = results[0].address_components;
            for (ac = 0; ac < arrAddress.length; ac++) {
                if (arrAddress[ac].types[0] == "street_number") { document.getElementById("tbUnit").value = arrAddress[ac].long_name }
                if (arrAddress[ac].types[0] == "route") { document.getElementById("tbStreet").value = arrAddress[ac].short_name }
                if (arrAddress[ac].types[0] == "locality") { document.getElementById("tbCity").value = arrAddress[ac].long_name }
                if (arrAddress[ac].types[0] == "administrative_area_level_1") { document.getElementById("tbState").value = arrAddress[ac].short_name }
                if (arrAddress[ac].types[0] == "postal_code") { document.getElementById("tbZip").value = arrAddress[ac].long_name }
            }
            document.getElementById("tbAddress").value = results[0].formatted_address;
        }
        document.getElementById("pResult").innerHTML = 'GeoCode Status:' + status;
    })
}
于 2014-02-13T02:42:54.580 回答
1

如果存在则返回位置。如果不是 - 返回administrative_area_1

city = results[0].address_components.filter(function(addr){
   return (addr.types[0]=='locality')?1:(addr.types[0]=='administrative_area_level_1')?1:0;
});
于 2016-11-21T19:38:35.967 回答
0 
            //if (arrAddress[ac].types[0] == "street_number") { alert(arrAddress[ac].long_name) } // SOKAK NO
            //if (arrAddress[ac].types[0] == "route") { alert(arrAddress[ac].short_name); } // CADDE
            //if (arrAddress[ac].types[0] == "locality") { alert(arrAddress[ac].long_name) } // İL
            //if (arrAddress[ac].types[0] == "administrative_area_level_1") { alert(arrAddress[ac].short_name) } // İL
            //if (arrAddress[ac].types[0] == "postal_code") { alert(arrAddress[ac].long_name); } // POSTA KODU
            //if (arrAddress[ac].types[0] == "neighborhood") { alert(arrAddress[ac].long_name); } // Mahalle
            //if (arrAddress[ac].types[0] == "sublocality") { alert(arrAddress[ac].long_name); } // İlçe
            //if (arrAddress[ac].types[0] == "country") { alert(arrAddress[ac].long_name); } // Ülke
于 2014-05-26T22:38:05.567 回答
0

以下是您可以与 lodash js 库一起使用的一些代码：（只需将 $scope.x 替换为您自己的变量名即可存储值）

    _.findKey(vObj.address_components, function(component) {

            if (component.types[0] == 'street_number') {
                $scope.eventDetail.location.address = component.short_name
            }

            if (component.types[0] == 'route') {
                $scope.eventDetail.location.address = $scope.eventDetail.location.address + " " + component.short_name;
            }

            if (component.types[0] == 'locality') {
                $scope.eventDetail.location.city = component.long_name;
            }

            if (component.types[0] == 'neighborhood') {
                $scope.eventDetail.location.neighborhood = component.long_name;
            }

        });
于 2015-05-04T20:33:42.490 回答
0

我使用了一个名为 find 的 lodash 函数，它返回谓词返回 true 的对象。就如此容易！

let city = find(result, (address) => {
  return typeof find(address.types, (a) => { return a === 'locality'; }) === 'string';
});
于 2016-03-17T22:46:22.683 回答
0

好吧，如果你想去这座城市，这对我有用

var city = "";
function getPositionByLatLang(lat, lng) {
    geocoder = new google.maps.Geocoder();
    var latlng = new google.maps.LatLng(lat, lng);
    geocoder.geocode({ 'latLng': latlng }, function (results, status) {
        if (status == google.maps.GeocoderStatus.OK) {
            if (results[1]) {
                //formatted address
                city = results[0].plus_code.compound_code;
                city = city.substr(0, city.indexOf(','));
                city = city.split(' ')[1];

                console.log("the name of city is: "+ city);
            }
        } else {
            // alert("Geocoder failed due to: " + status);
        }
    });
}
于 2020-06-03T10:54:56.267 回答
0

无需迭代即可获得城市，通常城市位于 address_components 对象的第二个键上，因此第二个表示 1：

results[0].address_components[1].long_name
于 2021-03-13T14:06:45.470 回答