Question

我编写了php代码来从数据库表生成xml。这个xml文件包含lat，lng的标记。然后我写了这段代码来显示地图上的标记：

<!DOCTYPE html >
<head>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />
<meta http-equiv="content-type" content="text/html; charset=UTF-8"/>
<title>PHP/MySQL & Google Maps Example</title>
<script type="text/javascript" src="https://maps.googleapis.com/maps/api/js"></script>
<script type="text/javascript">
//<![CDATA[
<?php
require("phpsqlsearch_genxml.php");
?>
var customIcons = {
  restaurant: {
    icon: 'http://labs.google.com/ridefinder/images/mm_20_blue.png'
  },
   bar: {
     icon: 'http://labs.google.com/ridefinder/images/mm_20_red.png'
   }
   };

  function load() {
   var map = new google.maps.Map(document.getElementById("map"), {
     center: new google.maps.LatLng(24.6557, 46.7351),
    zoom: 13,
    mapTypeId: 'roadmap'
     });
    var infoWindow = new google.maps.InfoWindow;

  // Change this depending on the name of your PHP file
  downloadUrl("test.xml", function(data) {
    var xml = data.responseXML;
    var markers = xml.documentElement.getElementsByTagName("marker");
    for (var i = 0; i < markers.length; i++) {
      var name = markers[i].getAttribute("name");
      var address = markers[i].getAttribute("address");
      //var type = markers[i].getAttribute("type");
      var point = new google.maps.LatLng(
          parseFloat(markers[i].getAttribute("lat")),
          parseFloat(markers[i].getAttribute("lng")));
      var html = "<b>" + name + "</b> <br/>" + address;
      var icon = customIcons[type] || {};
      var marker = new google.maps.Marker({
        map: map,
        position: point,
        icon: icon.icon
      });
      bindInfoWindow(marker, map, infoWindow, html);
    }
  });
}

function bindInfoWindow(marker, map, infoWindow, html) {
  google.maps.event.addListener(marker, 'click', function() {
    infoWindow.setContent(html);
    infoWindow.open(map, marker);
  });
}

function downloadUrl(url, callback) {
  var request = window.ActiveXObject ?
      new ActiveXObject('Microsoft.XMLHTTP') :
      new XMLHttpRequest;

  request.onreadystatechange = function() {
    if (request.readyState == 4) {
      request.onreadystatechange = doNothing;
      callback(request, request.status);
    }
  };

  request.open('GET', url, true);
  request.send(null);
}

function doNothing() {}

  //]]>

 </script>

 </head>

 <body onload="load()">
 <div id="map" style="width: 500px; height: 300px"></div>
 </body>
</html>

但是我得到了以下错误信息：

XML Parsing Error: junk after document element
Location: http://localhost/onlinetaxi/1.php
Line Number 8, Column 5:    var customIcons = {
----^

知道文件phpsqlsearch_genxml.php生成xml文件并将其保存为“test.xml”在同一目录中。任何人都可以告诉我为什么以及如何修复它??? 如果我删除了这些行：

网页将显示地图，但没有标记。为什么?????

在地图上显示mysql表中的多个标记

0 个答案: