如何从Google Maps API-地方获取完整地址?

时间:2019-01-30 00:34:19

标签: javascript google-maps-api-3

我正在尝试从Google地图自动填充中获取完整的地址,但是它不起作用。查看我的代码和控制台

var placeSearch, autocomplete;
var componentForm = [
    'delivery-street',
    'delivery-suburb',
    'delivery-postcode',
    'delivery-city',
    'delivery-country'
];

window.initAutocomplete = function() {
    // Create the autocomplete object, restricting the search to geographical
    // location types.
    autocomplete = new google.maps.places.Autocomplete(
        /** @type {!HTMLInputElement} */(document.getElementById('autocomplete')),
        {types: ['geocode']});

    // When the user selects an address from the dropdown, populate the address
    // fields in the form.
    autocomplete.addListener('place_changed', fillInAddress);
}

function fillInAddress() {
    // Get the place details from the autocomplete object.
    var place = autocomplete.getPlace();

    console.log(place);
    for (var component in componentForm) {
        document.getElementById(componentForm[component]).value = '';
        document.getElementById(componentForm[component]).disabled = false;
    }

    // Get each component of the address from the place details
    // and fill the corresponding field on the form.
    document.getElementById('delivery-street').value = findAttr(place, ['street_number']) + ' ' + findAttr(place, ['route']);
    document.getElementById('delivery-suburb').value = findAttr(place, ['sublocality_level_1', 'locality']);
    document.getElementById('delivery-postcode').value = findAttr(place, ['postal_code']);
    document.getElementById('delivery-city').value = findAttr(place, ['administrative_area_level_1', 'route']);
    document.getElementById('delivery-country').value = findAttr(place, ['country']);
}

function findAttr(place, findType) {
    var response = [];
    for (var j = 0; j < findType.length; j++) {

        for (var i = 0; i < place.address_components.length; i++) {
            var addressType = place.address_components[i].types[0];
            if(addressType === findType[j]){
                response.push(place.address_components[i].long_name);
            }
        }

    }
    return response.join(', ')
}

// Bias the autocomplete object to the user's geographical location,
// as supplied by the browser's 'navigator.geolocation' object.
window.geolocate = function () {
    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(function(position) {
            var geolocation = {
                lat: position.coords.latitude,
                lng: position.coords.longitude
            };
            var circle = new google.maps.Circle({
                center: geolocation,
                radius: position.coords.accuracy
            });
            autocomplete.setBounds(circle.getBounds());
        });
    }
}

因此,我输入了“ 2/89 Union St,New Brighton,Christchurch,New Zealand”,这就是控制台中的结果

address_components: Array(7)
0: {long_name: "89", short_name: "89", types: Array(1)}
1: {long_name: "Union Street", short_name: "Union St", types: Array(1)}
2: {long_name: "New Brighton", short_name: "New Brighton", types: Array(3)}
3: {long_name: "Christchurch", short_name: "Christchurch", types: Array(2)}
4: {long_name: "Canterbury", short_name: "Canterbury", types: Array(2)}
5: {long_name: "New Zealand", short_name: "NZ", types: Array(2)}
6: {long_name: "8061", short_name: "8061", types: Array(1)}

在文档中说

“自动完成”构造函数带有两个参数:

  • geocode指示Places服务仅返回地理编码 结果,而不是业务结果。
  • 地址指示Places服务仅返回地理编码结果 并提供准确的地址。

我尝试使用“地址”并且没有任何变化

  • 关于如何获取单位/单元/公寓编号的任何想法?

1 个答案:

答案 0 :(得分:0)

我遇到了同样的问题,最终只是创建了一个函数来构建所需的地址。这对美国非常有用。我不确定英国...

getAddressFromComponents(geo: any) {
    const streetNumber = geo.find( g => g.types.find( t => t === 'street_number') ).long_name;
    const streetName = geo.find( g => g.types.find( t => t === 'route' )).long_name;
    const cityName = geo.find( g => g.types.find( t => t === 'locality') && g.types.find( t => t === 'political' )).long_name;
    const stateName = geo.find( g => g.types.find( t => t === 'administrative_area_level_1') && g.types.find( t => t === 'political' ))
      .long_name;
    const countryName = geo.find( g => g.types.find( t => t === 'country') && g.types.find( t => t === 'political' )).long_name;
    const zip = geo.find( g => g.types.find( t => t === 'postal_code' )).long_name;

    return {
      address: `${streetNumber} ${streetName}`,
      city: cityName,
      state: stateName,
      country: countryName,
      postalCode: zip
    };
  }

用法

const address = this.getAddressFromComponents( _geoResponse[ 0 ].address_components );

出于价格上的考虑,我们将改用Algolia Places,因为它更易于使用:

https://community.algolia.com/places/