问题

Question

问题

我一直试图让this谷歌地图API示例正常工作，但在完全从谷歌的网站复制代码后，该示例无法正确呈现。即地图完全无法渲染，搜索栏为-350px的页面等。

要查看我的意思，请查看此JSfiddle。

该示例在Google上面提到的文档页面上运行得很好，所以我知道它正在做的事情。有什么想法吗？

编辑：我从谷歌的网站上获得了CSS，但问题仍然存在。

以下是导致问题的代码：

<!DOCTYPE html>
<html>
  <head>
    <title>Places Autocomplete</title>
    <meta name="viewport" content="initial-scale=1.0, user-scalable=no">
    <meta charset="utf-8">
    <link href="CSS/default.css" rel="stylesheet">
    <script src="https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false&libraries=places"></script>

    <style>
      input {
        border: 1px solid  rgba(0, 0, 0, 0.5);
      }
      input.notfound {
        border: 2px solid  rgba(255, 0, 0, 0.4);
      }
    </style>

    <script>
function initialize() {
  var mapOptions = {
    center: new google.maps.LatLng(-33.8688, 151.2195),
    zoom: 13,
    mapTypeId: google.maps.MapTypeId.ROADMAP
  };
  var map = new google.maps.Map(document.getElementById('map-canvas'),
    mapOptions);

  var input = /** @type {HTMLInputElement} */(document.getElementById('searchTextField'));
  var autocomplete = new google.maps.places.Autocomplete(input);

  autocomplete.bindTo('bounds', map);

  var infowindow = new google.maps.InfoWindow();
  var marker = new google.maps.Marker({
    map: map
  });

  google.maps.event.addListener(autocomplete, 'place_changed', function() {
    infowindow.close();
    marker.setVisible(false);
    input.className = '';
    var place = autocomplete.getPlace();
    if (!place.geometry) {
      // Inform the user that the place was not found and return.
      input.className = 'notfound';
      return;
    }

    // If the place has a geometry, then present it on a map.
    if (place.geometry.viewport) {
      map.fitBounds(place.geometry.viewport);
    } else {
      map.setCenter(place.geometry.location);
      map.setZoom(17);  // Why 17? Because it looks good.
    }
    marker.setIcon(/** @type {google.maps.Icon} */({
      url: place.icon,
      size: new google.maps.Size(71, 71),
      origin: new google.maps.Point(0, 0),
      anchor: new google.maps.Point(17, 34),
      scaledSize: new google.maps.Size(35, 35)
    }));
    marker.setPosition(place.geometry.location);
    marker.setVisible(true);

    var address = '';
    if (place.address_components) {
      address = [
        (place.address_components[0] && place.address_components[0].short_name || ''),
        (place.address_components[1] && place.address_components[1].short_name || ''),
        (place.address_components[2] && place.address_components[2].short_name || '')
      ].join(' ');
    }

    infowindow.setContent('<div><strong>' + place.name + '</strong><br>' + address);
    infowindow.open(map, marker);
  });

  // Sets a listener on a radio button to change the filter type on Places
  // Autocomplete.
  function setupClickListener(id, types) {
    var radioButton = document.getElementById(id);
    google.maps.event.addDomListener(radioButton, 'click', function() {
      autocomplete.setTypes(types);
    });
  }

  setupClickListener('changetype-all', []);
  setupClickListener('changetype-establishment', ['establishment']);
  setupClickListener('changetype-geocode', ['geocode']);
}

google.maps.event.addDomListener(window, 'load', initialize);

    </script>
  </head>
  <body>
    <div id="panel" style="margin-left: -260px">
      <input id="searchTextField" type="text" size="50">
      <input type="radio" name="type" id="changetype-all" checked="checked">
      <label for="changetype-all">All</label>

      <input type="radio" name="type" id="changetype-establishment">
      <label for="changetype-establishment">Establishments</label>

      <input type="radio" name="type" id="changetype-geocode">
      <label for="changetype-geocode">Geocodes</lable>
    </div>
    <div id="map-canvas"></div>
  </body>
</html>

Answer 1

JS字段不接受外部资源的Google Maps API库。（我不知道为什么）因此，您需要将＆lt; script＆gt;标记插入正文字段。

JS字段自动生成“window.onload”。您不必在JS字段中编写“google.maps.event.addDomListener”。

enter image description here http://jsfiddle.net/wf9a5m75/NpYwE/27/

<div id="panel" style="margin-left: -260px">
    <input id="searchTextField" type="text" size="50">
    <input type="radio" name="type" id="changetype-all" checked="checked">
    <label for="changetype-all">All</label>
    <input type="radio" name="type" id="changetype-establishment">
    <label for="changetype-establishment">Establishments</label>
    <input type="radio" name="type" id="changetype-geocode">
    <label for="changetype-geocode">Geocodes</lable>
</div>
<div id="map-canvas"></div>
        <script src="http://maps.google.com/maps/api/js?sensor=false&libraries=places"></script>

Answer 2

你会注意到javascript控制台中的错误：

Failed to load resource: the server responded with a status of 404 (Not Found) http://fiddle.jshell.net/maps/documentation/javascript/examples/default.css
Uncaught ReferenceError: google is not defined fiddle.jshell.net/agconti/NpYwE/show/:133

您没有在小提琴上正确包含Google API脚本。

Google Places API示例

问题

编辑：我从谷歌的网站上获得了CSS，但问题仍然存在。

2 个答案: