更改腳本的JavaScript

Question

我腳本calculate distance between 2 coordinates，在這裏：

(function() { 
     window.onload = function() { 
      var map; 

      //Parameter Google maps 
      var options = { 
       zoom: 10, //level zoom 
       //posisi tengah peta 
       center: new google.maps.LatLng(-7.695001, 113.996544), 
       mapTypeId: google.maps.MapTypeId.ROADMAP 
      }; 

      // Buat peta di menu MAP 
      var map = new google.maps.Map(document.getElementById('peta'), options); 
       var locations = [ 
       <?php while($rs = mysql_fetch_array($a_peta, MYSQL_ASSOC)) { ?> 
       ['<?php echo $rs['nama'] ?>', <?php echo $rs['latitude'] ?>, <?php echo $rs['longitude'] ?>], 
       <?php } ?> 
      ]; 

      var infowindow = new google.maps.InfoWindow(); 

      var marker, i; 
      /* kode untuk menampilkan banyak marker */ 
      for (i = 0; i < locations.length; i++) { 
       marker = new google.maps.Marker({ 
        position: new google.maps.LatLng(locations[i][1], locations[i][2]), 
        map: map, 
        fmap: fmap, 
        icon: 'pics/ico.jpg' 
       }); 

       var circle = new google.maps.Circle({ 
        map: map, 
        radius: 10000,  
        fillColor: '#AA0000' 
       }); 
       circle.bindTo('center', marker, 'position'); 
       /* menambahkan event clik untuk menampikan 
       infowindows dengan isi sesuai dengan 
       marker yang di klik */ 

       google.maps.event.addListener(marker, 'click', (function(marker, i) { 
        return function() { 
         infowindow.setContent(locations[i][0]); 
         infowindow.open(map, marker); 
        } 
       })(marker, i)); 
      } 
     }; 
    })(); 

你可以看到，代碼是php &的JavaScript之間的混合。我想將它移入一個完整的javascipt文件，並且不知道如何去做。

這裏是關於 '$ a_peta' 腳本：

<?php 
    include 'connect.php'; 

    $conn = mysql_connect($dbhost,$dbuser,$dbpass); 
    $db = mysql_select_db($dbname); 
    if($db){ 
     $sql="select id, nama, latitude, longitude from jarak order by id"; 
     $a_peta = mysql_query($sql); 
    } 
    else { 
     print "Database NOT Found "; 
     mysql_close($conn); 
    } 
?> 

Answer 1

嘗試使用AJAX調用，你可以在這裏看到的爲例： http://www.caveofprogramming.com/php/php-json-an-example-javascript-json-client-with-php-server/