Google Maps JavaScript Places API - URL undefined
我正在将Google Maps API集成到我正在处理的网站中。大多数情况下,它会按需要执行;但是,在拨打附近的搜索服务时,我很难获取返回的网址(位置的Google地图页面)。根据我对Google文档的理解,这个调用返回一个带有各种属性的PlaceResult对象,包括url.I能够正确访问其他两个属性,名称和附近,但是url属性未定义。这个问题可能是什么?谢谢。Google Maps JavaScript Places API - URL undefined
相关的代码片段:
var keyword = document.getElementById("searchBox").value;
var requestOptions = {
location: { lat: 37.3011339, lng: -89.5770238},
radius: '5000',
keyword: keyword
};
placesService = new google.maps.places.PlacesService(hotelMap);
placesService.nearbySearch(requestOptions, findCallback);
});
}; // end initiallize
function findCallback(results, status) {
var resultsList = "";
if (status == google.maps.places.PlacesServiceStatus.OK) {
for (var i = 0; i < results.length; i++) {
alert(results[i].url); // this returns undefined
resultsList += "<li>" + results[i].name + ": " + results[i].vicinity + " - <a href='" + results[i].url + "'>View Details</a></li>";
}
document.getElementById("searchList").innerHTML = resultsList;
}
}
参考:
https://developers.google.com/maps/documentation/javascript/reference#PlaceResult
看起来,物业是不是在nearbySearch
查询返回的PlaceResult
可用。该查询返回的结果中的placeIds详细信息请求会返回包含该属性的结果。
根据我对文档的阅读,唯一应该返回“受限制”PlaceResult
的查询是radarSearch
,但这似乎不是它工作的方式。
代码片段:
var geocoder;
var hotelMap;
var infowindow = new google.maps.InfoWindow();
function initialize() {
hotelMap = new google.maps.Map(
document.getElementById("map_canvas"), {
center: new google.maps.LatLng(37.4419, -122.1419),
zoom: 13,
mapTypeId: google.maps.MapTypeId.ROADMAP
});
var keyword = document.getElementById("searchBox").value;
var requestOptions = {
location: {
lat: 37.3011339,
lng: -89.5770238
},
radius: '5000',
keyword: keyword
};
placesService = new google.maps.places.PlacesService(hotelMap);
placesService.nearbySearch(requestOptions, findCallback);
}; // end initiallize
function findCallback(results, status) {
var resultsList = "";
if (status == google.maps.places.PlacesServiceStatus.OK) {
var bounds = new google.maps.LatLngBounds();
for (var i = 0; i < results.length; i++) {
console.log(results[i].url);
console.log(results[i]);
resultsList += "<li>" + results[i].name + ": " + results[i].vicinity + " - <a href='" + results[i].url + "'>View Details</a></li>";
var marker = new google.maps.Marker({
position: results[i].geometry.location,
map: hotelMap,
title: results[i].name,
placeId: results[i].place_id
});
bounds.extend(marker.getPosition());
google.maps.event.addListener(marker, 'click', (function(marker) {
return function(evt) {
var service = new google.maps.places.PlacesService(hotelMap);
service.getDetails({
placeId: this.placeId
}, function(place, status) {
if (status === google.maps.places.PlacesServiceStatus.OK) {
var content = '<div><strong>' + place.name + '</strong><br>' +
'Place ID: ' + place.place_id + '<br>' +
place.formatted_address + '<br>';
if (place.url) {
content += '<a href=' + place.url + ' target="_blank">[link]</a>';
}
content += '</div>';
infowindow.setContent(content);
infowindow.setPosition(place.geometry.location);
infowindow.open(hotelMap, marker);
}
});
}
})(marker));
}
hotelMap.fitBounds(bounds);
}
document.getElementById("searchList").innerHTML = resultsList;
}
google.maps.event.addDomListener(window, "load", initialize);
html,
body,
#map_canvas {
height: 100%;
width: 100%;
margin: 0px;
padding: 0px
}
<script src="https://maps.googleapis.com/maps/api/js?libraries=places"></script>
<input id="searchBox" value="coffee" />
<div id="map_canvas"></div>
<div id="searchList"></div>
非常感谢您的深入了解。我希望它不是这样的(就是说,nearSearch返回一个带有url属性的PlacesResult),但显然我们可以做的并不多。再次感谢您的发现。 – KellyMarchewa
测试在更多的地方?你正在看的地方可能实际上并没有谷歌网页?或者,使用网站对象,只显示可用。 –