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我有脚本来计算 2 个坐标之间的距离,这里:

(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));
            }
        };
    })();

您可以看到代码是 php 和 javascript 之间的混合。我想将它移动到一个完整的 javascipt 文件中,但不知道该怎么做。

这是关于 '$a_peta' 的脚本:

<?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);
    }
?>
4

1 回答 1

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尝试使用 Ajax 调用,您可以在这里看到一个示例:http: //www.caveofprogramming.com/php/php-json-an-example-javascript-json-client-with-php-server/

于 2013-05-29T09:57:28.347 回答