javascript - 标记未显示在谷歌地图上，但控制台中没有错误

Question

我已经为此纠结了一段时间:尝试使用 json 从我的数据库中获取 lat/lng 值，然后使用这些值将标记输出到 google map 上。目前 map 显示时没有标记，控制台中也没有错误。任何帮助将不胜感激!

 <script type="text/javascript">
       var map = null;

       function addMarker(lat,lng) {
             marker = new google.maps.Marker({
             position: new google.maps.LatLng(lat,lng),
             map: map,
           });
     }


      function initialize() {
      var mapOptions = {
      center: {lat: 54.872128, lng: - 6.284874},
      zoom: 13
     };
    map = new google.maps.Map(document.getElementById('map-canvas'),
    mapOptions);

   $(document).ready(function(){

   $.getJSON('MarkersController.php', function(data){


         var locations = JSON.parse(data);

         for(var i =0; i<locations.length; i++){
               addMarker(locations['lat'], locations['lng']);
      }
});
});


}

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

用于从 mysql 检索的 php 脚本:

<?php

include('DB_INFO.php');

// Opens a connection to a MySQL server.

$connection = mysqli_connect($server, $username, $password);

if (!$connection) {
    die('Not connected : ' . mysqli_error());
}
// Sets the active MySQL database.
$db_selected = mysqli_select_db($database, $connection);

if (!$db_selected) {
    die('Can\'t use db : ' . mysqli_error());
}

// Selects all the rows in the markers table.
$query = 'SELECT * FROM tester';
$result = mysqli_query($query);

if (!$result) {
    die('Invalid query: '. mysqli_error());
}

$list = array();
while ($row = mysqli_fetch_assoc($result)) {
    $list[] = $row;
}
header('Content-Type: application/json');
echo json_encode($list);
?>

Answer 1

如果你有一个标记集合，我认为你应该使用这种方式迭代集合并分配 i 元素:

for(var i =0; i<locations.length; i++){
  addMarker(locations[i].lat, locations[i].lng);
}

proof of concept fiddle