通过页面将变量传递到脚本中

Passing Variable Through a Page into a script

好的,这是我的问题。我有几页在地图上显示点。这些点对应于为我的员工完成的工作上传的照片数据库中的纬度和经度。 我只想显示与特定工单对应的点。这就是我所拥有的。 这是我的 Index.php 文件

<?php include 'active.php';
$work_order = $_REQUEST['work_order'];
global $work_order;
?>

<!DOCTYPE html>

<html lang="en">
<head>
    <meta charset="utf-8" />
    <title>Work Order GPS Data</title>
    <style type="text/css">
        body { font: normal 14px Verdana; }
        h1 { font-size: 24px; }
        h2 { font-size: 18px; }
        #sidebar { float: right; width: 30%; }
        #main { padding-right: 15px; }
        .infoWindow { width: 220px; }
    </style>
    <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script>
    <script type="text/javascript">
    //<![CDATA[
function makeRequest(url, callback) {
var request;
if (window.XMLHttpRequest) {
    request = new XMLHttpRequest(); // IE7+, Firefox, Chrome, Opera, Safari
} else {
    request = new ActiveXObject("Microsoft.XMLHTTP"); // IE6, IE5
}
request.onreadystatechange = function() {
    if (request.readyState == 4 && request.status == 200) {
        callback(request);
    }
}
request.open("GET", url, true);
request.send();
}     
    var map;

    var center = new google.maps.LatLng(38.988297, -75.785499);
    var geocoder = new google.maps.Geocoder();
var infowindow = new google.maps.InfoWindow();

function init() {

var mapOptions = {
  zoom: 9,
  center: center,
  mapTypeId: google.maps.MapTypeId.ROADMAP
}

map = new google.maps.Map(document.getElementById("map_canvas"), mapOptions);

makeRequest('get_locations.php?work_order=$work_order', function(data) {

    var data = JSON.parse(data.responseText);

    for (var i = 0; i < data.length; i++) {
        displayLocation(data[i]);
    }
});
}

function displayLocation(location) {

var content =   '<div class="infoWindow"><strong>'  + location.name +     '</strong>'
                + '<br/>'     + location.date_time
                + '<br/>'     + location.work_order + '</div>';

if (parseInt(location.lat) == 0) {
    geocoder.geocode( { 'address': location.address }, function(results, status) {
        if (status == google.maps.GeocoderStatus.OK) {

            var marker = new google.maps.Marker({
                map: map, 
                position: results[0].geometry.location,
                title: location.name
            });

            google.maps.event.addListener(marker, 'click', function() {
                infowindow.setContent(content);
                infowindow.open(map,marker);
            });
        }
    });
} else {
    var position = new google.maps.LatLng(parseFloat(location.lat), parseFloat(location.lon));
    var marker = new google.maps.Marker({
        map: map, 
        position: position,
        title: location.name
    });

    google.maps.event.addListener(marker, 'click', function() {
        infowindow.setContent(content);
        infowindow.open(map,marker);
    });
}
}


    </script>
</head>
<center>
<body onload="init();">

    <h1>Work Order GPS Data</h1>

    <section id="sidebar">
        <div id="directions_panel"></div>
    </section>

    <section id="main">
        <div id="map_canvas" style="width: 95%; height: 840px;"></div>
    </section>

</body>
</center>
</html>

然后我有我的get_locations.php

<?php

require 'config.php';
$work_order = $_REQUEST['work_order'];

try {

    $db = new PDO($dsn, $username, $password);
    $db->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
    if($work_order != ''){ 
    $sth = $db->query("SELECT * FROM exif WHERE work_order=$work_order");
    }else{$sth = $db->query("SELECT * FROM exif");
    }
    $locations = $sth->fetchAll();

    echo json_encode( $locations );

} catch (Exception $e) {
    echo $e->getMessage();
}

我的问题是,当我去 http://work.newmansecurity.com/gps/?work_order=1234 时,我不仅获得工作订单 1234 的积分,我还获得了所有积分。但是,如果我直接转到 /gps/get_locations.php?work_order=1234 它会起作用。所以出于某种原因,变量没有按照我的意愿传递。

请帮助

谢谢

尝试用单引号将查询中的变量 $work_order 括起来。

*****已解决*****

更改了

function init() {

var mapOptions = {
  zoom: 9,
  center: center,
  mapTypeId: google.maps.MapTypeId.ROADMAP
}

map = new google.maps.Map(document.getElementById("map_canvas"), mapOptions);

makeRequest(<?php echo json_encode($url); ?>, function(data) {

    var data = JSON.parse(data.responseText);

    for (var i = 0; i < data.length; i++) {
        displayLocation(data[i]);
    }
});
}

对此。使用 json 编码传递变量。效果很好