在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，但不是item.address_components.locality？
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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直流
答案 0 :(得分：57)
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
});
答案 1 :(得分：15)
尝试了几个不同的请求：
就像你说的那样，返回的数组大小不一致，但两个结果的Town似乎都在address_component项中，类型为[“locality”，“political”]。也许你可以用它作为指标？
编辑：使用jQuery获取位置对象，将其添加到响应函数中：
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
});
答案 2 :(得分：10)
当用户点击地图上的某个位置时，我必须创建一个程序，用于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，是一个允许公众为数据库贡献的用户表单。我并不认为自己是专家，但效果很好。
<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>
答案 3 :(得分：6)
我假设您想要获得城市和州/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。您必须调用函数，例如updateAddress（）;
答案 4 :(得分：4)
我创建了此函数以获取地址解析器结果的主要信息：
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);
假设您要搜索Santiago Bernabéu Stadium，那么结果将是：
{
country: 'Spain',
locality: 'Madrid',
route: 'Avenida de Concha Espina',
streetNumber: '1',
}
答案 5 :(得分：3)
我认为谷歌并没有提供某种功能来实现这些目标，这真是一种痛苦。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);
}
});
答案 6 :(得分：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;
})
}
答案 7 :(得分：1)
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;
我正在Node中执行此操作，所以没关系。如果您需要支持IE，则需要为Array.prototype.includes使用polyfill或找到另一种方法。
答案 8 :(得分：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
答案 9 :(得分：0)
这里有一些可以与lodash js库一起使用的代码:(只需用你自己的变量名替换$ scope.x来存储值）
_.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;
}
});
答案 10 :(得分：0)
我使用了一个名为find的lodash函数，该函数返回谓词返回true的对象。就这么简单！
let city = find(result, (address) => {
return typeof find(address.types, (a) => { return a === 'locality'; }) === 'string';
});
答案 11 :(得分：0)
如果存在则返回locality。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;
});
答案 12 :(得分：0)
如果您想获得这座城市，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);
}
});
}
答案 13 :(得分：0)
不用迭代就可以得到城市，一般城市位于address_components对象的第二个键上，所以第二个是1：
results[0].address_components[1].long_name