Question

有谁知道如何使用此处提到的距离搜索选项排名？ https://developers.google.com/maps/documentation/javascript/places#place_search_requests

在请求选项中列出这个似乎不起作用。这是我相对于此的代码部分：

var request = {
  location: coords,
  //radius: 30000,
  keyword: ['puma, retail'],
  types: ['store'],
  rankBy: google.maps.places.RankBy.DISTANCE
};

service = new google.maps.places.PlacesService(map);
service.search(request, callback);

function callback(results, status) {
    if (status == google.maps.places.PlacesServiceStatus.OK) {
        for (var i = 0; i < results.length; i++) {
            createMarker(results[i]);
                            listResults(results[i]);
        }
    } 
}

如果我包含半径，代码将定位并列出结果，但结果不按距离按升序列出。谷歌的文档说，如果使用rankBy选项，则不需要半径。我错过了什么吗？

Answer 1

您不能同时使用radius和RankBy.DISTANCE属性。因此，您有两种选择：

1）按半径搜索，然后在您自己的代码中按距离对结果进行排序。

示例：

var request = {
               location: coords,
               radius: 30000,
               keyword: ['puma, retail'],
               types: ['store']
               };
service = new google.maps.places.PlacesService(map);
service.search(request, callback);

function callback(results, status) {
         if (status == google.maps.places.PlacesServiceStatus.OK) {
          for (var i = 0; i < results.length; i++) {
           sortresults(results[i]);//sortresult uses haversine to calcuate distance and then arranges the result in the order of distance
           createMarker(results[i]);
           listResults(results[i]);
       }
   } 
}

选项2：按RankBy.Distance搜索，然后使用半径限制结果。您需要再次使用半正式公式来计算距离。

var request = {
               location: coords,
               rankBy: google.maps.places.RankBy.DISTANCE,
               keyword: ['puma, retail'],
               types: ['store']
               };
service = new google.maps.places.PlacesService(map);
service.search(request, callback);

function callback(results, status) {
         if (status == google.maps.places.PlacesServiceStatus.OK) {

           for (var i = 0; i < results.length; i++) {
           d= distance(coords,results[i].latlng)
           if(d<rd)
            {createMarker(results[i]);
             listResults(results[i]);
            }
           }
   } 
}

//Returns Distance between two latlng objects using haversine formula
distance(p1, p2) {
 if (!p1 || !p2) 
  return 0;
 var R = 6371000; // Radius of the Earth in m
 var dLat = (p2.lat() - p1.lat()) * Math.PI / 180;
 var dLon = (p2.lng() - p1.lng()) * Math.PI / 180;
 var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
 Math.cos(p1.lat() * Math.PI / 180) * Math.cos(p2.lat() * Math.PI / 180) *
 Math.sin(dLon / 2) * Math.sin(dLon / 2);
 var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
 var d = R * c;
 return d;
 }

Answer 2

遇到了同样的问题。根据此来源：http://www.geocodezip.com/v3_GoogleEx_place-search.html 我能够如此制定查询：

var request = {
location: gps,
types: ['food'], //You can substitute "keyword: 'food'," (without double-quotes) here as well.
rankBy: google.maps.places.RankBy.DISTANCE, //Note there is no quotes here, I made that mistake.
key: key 
};

Var键是API键，它不是必需的，但添加后供使用。 GPS是：

var gps = new google.maps.LatLng(location.lat,location.lon);

最后，我做了你所做的一切，除了我添加了一个我不打算使用的地图边界。为此我做了：

var bounds = new google.maps.LatLngBounds();

Answer 3

您不能同时使用radius + rankBy属性。

如果您需要最近的位置，请选择一些类型，并设置rankBy proprety。

places.nearbySearch({
                        location: LatLng,
                        types: placeTypes,
                        rankBy: google.maps.places.RankBy.DISTANCE
                    },
                    function (results) {
                        // process the results, r[0] is the closest place
                    }
                );

你可以在这里设置像这样的地方类型

var placeTypes = [
'accounting',
'airport',
'amusement_park',
'aquarium',
'art_gallery',
'atm',
'bakery',
'bank',
'bar',
'beauty_salon',
'bicycle_store',
'book_store',
'bowling_alley',
'bus_station',
'cafe',
'campground',
'car_dealer',
'car_rental',
'car_repair',
'car_wash',
'casino',
'cemetery',
'church',
'city_hall',
'clothing_store',
'convenience_store',
'courthouse',
'dentist',
'department_store',
'doctor',
'electrician',
'electronics_store',
'embassy',
'establishment',
'finance',
'fire_station',
'florist',
'food',
'funeral_home',
'furniture_store',
'gas_station',
'general_contractor',
'grocery_or_supermarket',
'gym',
'hair_care',
'hardware_store',
'health',
'hindu_temple',
'home_goods_store',
'hospital',
'insurance_agency',
'jewelry_store',
'laundry',
'lawyer',
'library',
'liquor_store',
'local_government_office',
'locksmith',
'lodging',
'meal_delivery',
'meal_takeaway',
'mosque',
'movie_rental',
'movie_theater',
'moving_company',
'museum',
'night_club',
'painter',
'park',
'parking',
'pet_store',
'pharmacy',
'physiotherapist',
'place_of_worship',
'plumber',
'police',
'post_office',
'real_estate_agency',
'restaurant',
'roofing_contractor',
'rv_park',
'school',
'shoe_store',
'shopping_mall',
'spa',
'stadium',
'storage',
'store',
'subway_station',
'synagogue',
'taxi_stand',
'train_station',
'travel_agency',
'university',
'veterinary_care',
'zoo'

];

Answer 4

请注意，在当前的Api Doc上，＆＃34;搜索＆＃34; PlacesService类上的方法不可用。

https://developers.google.com/maps/documentation/javascript/reference?hl=it#PlacesService

你必须选择下：

nearbySearch（每页检索20个结果，最多3页）
radarSearch（检索200结果但细节较少）
textSearch（类似于NearbySearch）

如果您选择RankBy.DISTANCE，则无法设置半径

var request = {
  location: vLatLng,
  //radius: vRaggio,
  rankBy: google.maps.places.RankBy.DISTANCE,
  keyword: ['puma, retail'],
  types: ['store']
}

placesService.nearbySearch(request, function (data, status, placeSearchPagination) {
  if (status == google.maps.places.PlacesServiceStatus.OK) {
  //...
  // do your stuffs with data results
  //...
  if (placeSearchPagination && placeSearchPagination.hasNextPage) {
    placeSearchPagination.nextPage();
  } 
});

如果您想根据距离半径限制数据，可以使用此功能检查重新开始是否太远。

function checkRadiusDistance(place,centerLatLng,radius) {
  return google.maps.geometry.spherical.computeDistanceBetween(place.geometry.location, centerLatLng) < radius;
});

请注意，当您指定＆＃34; rankBy：google.maps.places.RankBy.PROMINENCE时，这也是获取特定半径内的地点的唯一方法 EM>＆＃34;和＆＃34; radius = xx ＆＃34; placesService为您提供在定义区域之外的结果。

google maps api v3如何按最近距离排名

4 个答案: