发送用户Lat/Lng到php代码

所以这是我的困境，我希望能够使用用户的当前位置来填充索引页上的表，其中包含用户附近的位置。最好的方法是什么?以下是我用来收集用户的lat/lng值的代码:

<script type="text/javascript">
var x = document.getElementById("demo");
function getLocation() {
    if (navigator.geolocation) {
        navigator.geolocation.getCurrentPosition(showPosition, showError);
    } else { 
        x.innerHTML = "Geolocation is not supported by this browser.";
    }
}
function showPosition(position) {
    x.innerHTML = "Latitude: " + position.coords.latitude + 
    "<br>Longitude: " + position.coords.longitude;
}
function showError(error) {
    switch(error.code) {
        case error.PERMISSION_DENIED:
            x.innerHTML = "User denied the request for Geolocation."
            break;
        case error.POSITION_UNAVAILABLE:
            x.innerHTML = "Location information is unavailable."
            break;
        case error.TIMEOUT:
            x.innerHTML = "The request to get user location timed out."
            break;
        case error.UNKNOWN_ERROR:
            x.innerHTML = "An unknown error occurred."
            break;
    }
}
navigator.geolocation.getCurrentPosition(success, error);
    function success(position) {
        var GEOCODING = 'https://maps.googleapis.com/maps/api/geocode/json?latlng=' + position.coords.latitude + '%2C' + position.coords.longitude + '&language=en';
        $.getJSON(GEOCODING).done(function(location) {
             $('#latitude').html(position.coords.latitude);
             $('#longitude').html(position.coords.longitude);
               $.ajax({
                   url:'table.php',
                   method: 'POST',
                   dataType: 'json',
                   data:{
                       lat: $('#latitude').val(),
                       lng: $('#longitude').val()
                   },
                   success: function(data){
                       console.log();
                   }
                });             
             }
             function error(err) {
                 console.log(err)
             }
</script>

<input type="hidden" id="latitude">
<input type="hidden" id="longitude">

但是我如何将这些位置放入PHP变量并将它们传递给数据库查询，以便我可以在用户登陆时在索引页上显示它们

  <link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.12/css/jquery.dataTables.css">
        <script>
        $(document).ready(function(){
            var dataTable = $('#searchTable').dataTable();
            $("#searchbox").keyup(function() {
                dataTable.fnFilter(this.value);
            });
        });  
        </script>
        echo'<table id="searchTable" class="display">
                <thead>
                    <tr>
                        <th>Latitude</th>
                        <th>Longitude</th>
                    </tr>
                </thead>
                <tbody>
                ';
                (data to display from tables.php)
        echo'</tbody>
            </table>
                ';

(我使用数据表来显示位置)我应该通过AJAX将值发送到具有查询的单独PHP页面，然后再返回到主页面吗?这样，只要用户登陆到主页，查询就会更新为用户位置?非常感谢任何帮助。几天来，我一直在寻找解决办法，但运气不太好。谢谢。

这是我的表。php页面(仍在试图找出如何正确地发送数据回索引页)

<?php
require_once("connect.php");
$lat = isset($_POST['lat']) ? $_POST['lat'] : null;
$lng = isset($_POST['lng']) ? $_POST['lng'] : null;
if($lat && $lng){
    $locations = array();       
    $stmt = "SELECT * FROM location LIMIT 500"; //Query data from table
    $stmt = $conn->prepare($stmt);
    $stmt->execute();
    if($stmt->rowCount() > 0){
        $result = $stmt->fetchAll();
        foreach($result as $row){ 
             echo'<tr>
                <td>';              
                   $locations['Latitude'][] = $row;
                echo'</td>
                <td>';
                   $locations['Longitude'][] = $row;    
                echo'</td>
             </tr>';        
        }
    }
     return json_encode($locations);
}
?>