从地理编码结果中获取城市？

Question

从Geocoder结果获取不同数组内容时出现问题。

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

Answer 1

得到使用，到底这个工作：

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 
}); 

Answer 2

尝试了不同的要求：

MK107BX

Cleveland Park Crescent, UK

像你说的，返回数组的大小不一致，但对于两个结果城市似乎是在address_component项目，类型为[“locality”，“political”]。也许你可以用它作为一个指标？

编辑：使用jQuery获得当地对象，将它添加到您的响应功能：

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 
    }); 

Answer 3

我不得不创建程序当用户点击地图上的位置，将填写纬度，经度，市，县，和州字段用户的形式。该页面可以在http://krcproject.groups.et.byu.net发现，它是用户的形式，让公众作出贡献的数据库。我并不声称自己是专家，但它很棒。

<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> 

Answer 4

我假设你想要得到的城市，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 
} 

不要试图返回结果，你会发现他们是不确定的 - 的结果异步服务。你必须调用一个函数，比如updateAddress（）;

Answer 5

// 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; 
    }) 
} 

Answer 6

  //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 

Answer 7

我认为这是一个真正的痛苦，谷歌并没有提供某种功能，以得到这些。无论如何，我认为找到合适的对象的最好的办法是：

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); 
    }    
}); 

Answer 8

下面是一些代码，你可以用lodash JS库使用：（只是用自己的变量名替换$ scope.x存储的值）

_.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; 
      } 

     }); 

Answer 9

我用了一个叫做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'; 
}); 

Answer 10

返回locality是否存在。如果不是 - 退货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; 
});