Question

每当googlemaps视口中心发生变化时，我都会尝试使用邮政编码向我的数据库提交查询。我知道这可以通过反向地理编码完成，例如：

google.maps.event.addListener(map, 'center_changed', function(){
newCenter();
});
...
function newCenter(){
var newc = map.getCenter();
geocoder.geocode({'latLng': newc}, function(results, status){
if (status == google.maps.GeocoderStatus.OK) {
  var newzip = results[0].address_components['postal_code'];
  }
});
};

当然，这段代码实际上并不起作用。所以我想知道如何更改这个以便从结果数组中提取邮政编码。感谢

Answer 1

到目前为止我已经意识到在大多数情况下 ZIPCODE 总是每个返回地址中的最后一个值，所以，如果你想要检索它第一个zipcode（这是我的情况），您可以使用以下方法：

var address = results[0].address_components;
var zipcode = address[address.length - 1].long_name;

Answer 2

使用JQuery？

var searchAddressComponents = results[0].address_components,
    searchPostalCode="";

$.each(searchAddressComponents, function(){
    if(this.types[0]=="postal_code"){
        searchPostalCode=this.short_name;
    }
});

short_name或long_name将在上面工作 “searchPostalCode”var将包含邮政（zip？）代码IF 并且只有从Google Maps API获得一个有时你不会得到“postal_code”来换取你的查询。

Answer 3

您可以使用underscore.js库轻松完成此操作：http://documentcloud.github.com/underscore/#find

_.find(results[0].address_components, function (ac) { return ac.types[0] == 'postal_code' }).short_name

Answer 4

好吧，所以我明白了。解决方案比我想要的更加丑陋，我可能不需要最后一个for循环，但这里是需要从address_components []中提取废话的其他人的代码。这是在地理编码器回调函数

中

for(i; i < results.length; i++){
            for(var j=0;j < results[i].address_components.length; j++){
                for(var k=0; k < results[i].address_components[j].types.length; k++){
                    if(results[i].address_components[j].types[k] == "postal_code"){
                        zipcode = results[i].address_components[j].short_name;
                    }
                }
            }
    }

Answer 5

这只需要两个for循环。一旦我们发现第一个“类型”为“postal_code”，“results”数组就会更新。

然后用新发现的数组设置更新原始数组并再次循环。

            var i, j,
            result, types;

            // Loop through the Geocoder result set. Note that the results
            // array will change as this loop can self iterate.
            for (i = 0; i < results.length; i++) {

                result = results[i];

                types = result.types;

                for (j = 0; j < types.length; j++) {

                    if (types[j] === 'postal_code') {

                        // If we haven't found the "long_name" property,
                        // then we need to take this object and iterate through
                        // it again by setting it to our master loops array and 
                        // setting the index to -1
                        if (result.long_name === undefined) {
                            results = result.address_components;
                            i = -1;
                        }
                        // We've found it!
                        else {
                            postcode = result.long_name;
                        }

                        break;

                    }

                }

            }

Answer 6

$.each(results[0].address_components,function(index,value){
    if(value.types[0] === "postal_code"){
        $('#postal_code').val(value.long_name);
    }
});

Answer 7

您也可以使用此代码，此功能有助于在按钮点击或onblur或keyup或keydown上获取zip。

只需将地址传递给此功能。

使用谷歌api删除有效的密钥和传感器选项，因为它现在不需要。

$(document).ready(function() {
  var number = parseInt($('#director-uploads1').data('number'));

  function clone(){
    var director = $('.director-uploads-hidden').clone(true); //true will cause it to clone all associated data and events as well
    director.data("number", number+1);
    director.removeClass("director-uploads-hidden");
    director.css("display", "block"); 
    //if you want to add any other extra data or events, do it here before appending the object
    $('#director-uploads1').append(director);
    number++;
  }

  $("button#add-director").on("click", clone);
});

Answer 8

您还可以使用JavaScript .find方法，该方法类似于下划线_.find方法，但是它是本机的，不需要任何额外的依赖关系。

const zip_code = results[0].address_components.find(addr => addr.types[0] === "postal_code").short_name;

Answer 9

places.getDetails( request_details, function(results_details, status){

                // Check if the Service is OK
                if (status == google.maps.places.PlacesServiceStatus.OK) {                  

                    places_postal           = results_details.address_components
                    places_phone            = results_details.formatted_phone_number
                    places_phone_int        = results_details.international_phone_number
                    places_format_address   = results_details.formatted_address
                    places_google_url       = results_details.url
                    places_website          = results_details.website
                    places_rating           = results_details.rating

                    for (var i = 0; i < places_postal.length; i++ ) {
                        if (places_postal[i].types == "postal_code"){
                            console.log(places_postal[i].long_name)
                        }
                    }

                }
            });

这对我来说似乎非常有效，这与新的Google Maps API V3有关。如果这有助于任何人，请写评论，我正在编写我的脚本，因为它们可能会改变。

Answer 10

我使用此代码获取＆＃34;邮政编码＆＃34;和＆＃34; locality＆＃34;，但您可以使用它来获取任何其他字段只是更改类型的值：

<强> JAVASCRIPT

var address = results[0].address_components;
var zipcode = '';
var locality = '';

for (var i = 0; i < address.length; i++) {
     if (address[i].types.includes("postal_code")){ zipcode = address[i].short_name; }    
     if (address[i].types.includes("locality")){ locality = address[i].short_name; }
}

Answer 11

在PHP中我使用此代码。几乎在每种情况下它都有效。

$zip = $data["results"][3]["address_components"];
$zip = $index[0]["short_name"];

Answer 12

使用JSONPath，可以通过一行代码轻松完成：

var zip = $.results[0].address_components[?(@.types=="postal_code")].long_name;

Answer 13

我知道zip是最后一个，或者是最后一个。那就是为什么这是我的解决方法

const getZip = function (arr) {
  return (arr[arr.length - 1].types[0] === 'postal_code') ? arr[arr.length - 1].long_name : arr[arr.length - 2].long_name;
};
const zip = getZip(place.address_components);

Answer 14

我认为与其检查索引，不如更好地检查组件内部的地址类型键。我通过使用开关盒解决了这个问题。

      var address = '';
      var pin = '';
      var country = '';
      var state = '';
      var city = '';
      var streetNumber = '';
      var route ='';
      var place = autocomplete.getPlace(); 
      for (var i = 0; i < place.address_components.length; i++) {
        var component = place.address_components[i];
        var addressType = component.types[0];

        switch (addressType) {
            case 'street_number':
                streetNumber = component.long_name;
                break;
            case 'route':
                route = component.short_name;
                break;
            case 'locality':
                city = component.long_name;
                break;
            case 'administrative_area_level_1':
                state = component.long_name;
                break;
            case 'postal_code':
                pin = component.long_name;
                break;
            case 'country':
                country = component.long_name;
                break;
        }
       }

Answer 15

Romaine M. - 谢谢！如果您只需要在Google首次返回的结果中找到邮政编码，则只需执行2次循环：

for(var j=0;j < results[0].address_components.length; j++){
    for(var k=0; k < results[0].address_components[j].types.length; k++){
        if(results[0].address_components[j].types[k] == "postal_code"){
            zipcode = results[0].address_components[j].long_name;
        }
    }
}

Answer 16

现在似乎最好从宁静的API中获取它，只需尝试：

https://maps.googleapis.com/maps/api/geocode/json?latlng=40.714224,-73.961452&key=YOUR_KEY_HERE

使用AJAX GET调用非常有效！

类似的东西：

var your_api_key = "***";
var f_center_lat = 40.714224;
var f_center_lon = -73.961452;

$.ajax({ url: "https://maps.googleapis.com/maps/api/geocode/json?latlng="+f_center_lat+","+f_center_lon+"&key="+your_api_key,
            method: "GET"
        })
        .done(function( res ) { if (debug) console.log("Ajax result:"); console.log(res);
            var zipCode = null;
            var addressComponent = res.results[0].address_components;
            for (var x = 0 ; x < addressComponent.length; x++) {
                var chk = addressComponent[x];
                if (chk.types[0] == 'postal_code') {
                    zipCode = chk.long_name;
                }
            }
            if (zipCode) {
                //alert(zipCode);
                $(current_map_form + " #postalcode").val(zipCode);
            }
            else {
                //alert('No result found!!');
                if (debug) console.log("Zip/postal code not found for this map location.")
            }
        })
        .fail(function( jqXHR, textStatus ) {
            console.log( "Request failed (get postal code via geocoder rest api). Msg: " + textStatus );
        });

Answer 17

//autocomplete is the text box where u will get the suggestions for an address.

autocomplete.addListener('place_changed', function () {
//Place will get the selected place geocode and returns with the address              
//and marker information.
        var place = autocomplete.getPlace();
//To select just the zip code of complete address from marker, below loop //will help to find. Instead of y.long_name you can also use y.short_name. 
        var zipCode = null;
        for (var x = 0 ; x < place.address_components.length; x++) {
            var y = place.address_components[x];
            if (y.types[0] == 'postal_code') {
                zipCode = y.long_name;
            }
        }
    });

Answer 18

 return $http.get('//maps.googleapis.com/maps/api/geocode/json', {
            params: {
                address: val,
                sensor: false
            }
        }).then(function (response) {
           var model= response.data.results.map(function (item) {
               // return item.address_components[0].short_name;
               var short_name;
             var st=  $.each(item.address_components, function (value, key) {
                    if (key.types[0] == "postal_code") {
                        short_name= key.short_name;
                    }
                });
             return short_name;

            });
                return model;
        });

Answer 19

使用Jquery

您无法确定邮政编码存储在address_components数组中的哪个位置。有时在address_components.length - 1＆gt;密码可能不存在。在“地址到latlng”地理编码中也是如此。
您可以确定Postal代码将包含“postal_code”字符串。所以最好的方法就是检查一下。

   var postalObject = $.grep(results[0].address_components, function(n, i) {
                                    if (n.types[0] == "postal_code") {
                                        return n;
                                    } else {
                                        return null;
                                    }
                                });

                                $scope.query.Pincode = postalObject[0].long_name;

Answer 20

这个简单的代码适合我

for (var i = 0; i < address.length; i++) {
        alert(address[i].types);
        if (address[i].types == "postal_code")
            $('#postalCode').val(address[i].long_name);
        if (address[i].types == "")
            $('#country').val(address[i].short_name);
    }

Answer 21

总之，这是一项很大的努力。至少使用v2 API，我可以检索这些细节：

var place = response.Placemark[0];
var point = new GLatLng(place.Point.coordinates[1], place.Point.coordinates[0]);

myAddress = place.AddressDetails.Country.AdministrativeArea.SubAdministrativeArea.Locality.Thoroughfare.ThoroughfareName

myCity = place.AddressDetails.Country.AdministrativeArea.SubAdministrativeArea.Locality.LocalityName

myState = place.AddressDetails.Country.AdministrativeArea.AdministrativeAreaName

myZipCode = place.AddressDetails.Country.AdministrativeArea.SubAdministrativeArea.Locality.PostalCode.PostalCodeNumber

必须有一种更优雅的方式来检索单个address_components，而无需通过刚刚经历的循环jujitsu。

使用Google Maps Javascript API V3反向地理编码检索邮政编码

21 个答案: