如何从Google Places API获取原始地点评分？

Question

我正试图从Google地方信息中获得地点评分，但我坚持了这部分代码：

function initMap() {
  var map = new google.maps.Map();
  var service = new google.maps.places.PlacesService(map);

  service.getDetails({
      placeId: "ChIJ2fNZsMQ0GkcRCt39huGpijo"
    },
    function(place, status) {
      if (status === google.maps.places.PlacesServiceStatus.OK) {
        var rating = place.rating;
        $('.google_rating').html(rating);
      }
    });
}

<div class="rating" id="google_rating">
  <div class="google_rating"></div><span class="rating_star">&#9733;</span>
</div>

<script src="https://maps.googleapis.com/maps/api/js?key=AIzaSyB1awNWLs_8JOTHWaC08TCUOo0r0PxPqvY&libraries=places&callback=initMap" async defer></script>

我不需要任何地图，仅需要特定地点的原始评分（地点ID）。我想念什么？

Answer 1

PlacesService构造函数采用google.maps.Map对象或HTMLDivElement：

  

PlacesService（attrContainer）

     

参数：
  attrContainer：HTMLDivElement |地图
  创建PlacesService的新实例，以在指定容器中呈现归因。

如果您不需要地图，则不要用地图来构造它，请使用HTMLDivElement像这样：

var service = new google.maps.places.PlacesService(document.getElementById("placeAttr"));

代码段：

function initMap() {
  var service = new google.maps.places.PlacesService(document.getElementById("placeAttr"));

  service.getDetails({
      placeId: "ChIJ2fNZsMQ0GkcRCt39huGpijo"
    },
    function(place, status) {
      if (status === google.maps.places.PlacesServiceStatus.OK) {
        var rating = place.rating;
        $('.google_rating').html(rating);
      }
    });
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="rating" id="google_rating">
  <div class="google_rating"></div><span class="rating_star">&#9733;</span>
</div>
<div id="placeAttr"></div>
<script src="https://maps.googleapis.com/maps/api/js?key=AIzaSyB1awNWLs_8JOTHWaC08TCUOo0r0PxPqvY&libraries=places&callback=initMap" async defer></script>