Question

<html xmlns="http://www.w3.org/1999/xhtml">
  <head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8"/>
    <title>Google Maps AJAX + mySQL/PHP</title>
    <script src="http://maps.google.com/maps/api/js?sensor=false" type="text/javascript"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
    <script type="text/javascript">

    var customIcons = {
      restaurant: {
        icon: 'http://labs.google.com/ridefinder/images/mm_20_blue.png',
        shadow: 'http://labs.google.com/ridefinder/images/mm_20_shadow.png'
      },
      bar: {
        icon: 'http://labs.google.com/ridefinder/images/mm_20_red.png',
        shadow: 'http://labs.google.com/ridefinder/images/mm_20_shadow.png'
      }
    };
    function load() {
      var map = new google.maps.Map(document.getElementById("map"), {
        center: new google.maps.LatLng(47.6145, -122.3418),
        zoom: 13,
        mapTypeId: 'roadmap'
      });
      var infoWindow = new google.maps.InfoWindow;
      // Change this depending on the name of your PHP file
      downloadUrl("maps_xml.php", 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,
            shadow: icon.shadow
          });
          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: 100%; height: 100%"></div>
  </body>
</html>1

maps_xml.php 这是我的phpfile，它从数据库中获取数据并动态创建标记。但这样做我是geeting地图但我能够在地图上获得任何标记，即使插入了大约7个数据。所以请帮助我解决这个问题。

<?php
require("config.php");
function parseToXML($htmlStr) 
{ 
$xmlStr=str_replace('<','&lt;',$htmlStr); 
$xmlStr=str_replace('>','&gt;',$xmlStr); 
$xmlStr=str_replace('"','&quot;',$xmlStr); 
$xmlStr=str_replace("'",'&apos;',$xmlStr); 
$xmlStr=str_replace("&",'&amp;',$xmlStr); 
return $xmlStr; 
} 
// Opens a connection to a mySQL server
$connection=mysql_connect (localhost,$username, $password);
if (!$connection) {
  die('Not connected : ' . mysql_error());
}
// Set the active mySQL database
$db_selected = mysql_select_db($database, $connection);
if (!$db_selected) {
  die ('Can\'t use db : ' . mysql_error());
}
// Select all the rows in the markers table
$query = "SELECT * FROM markers WHERE 1";
$result = mysql_query($query);
if (!$result) {
  die('Invalid query: ' . mysql_error());
}
header("Content-type: text/xml");
// Start XML file, echo parent node
echo '<markers>';
// Iterate through the rows, printing XML nodes for each
while ($row = @mysql_fetch_assoc($result)){
  // ADD TO XML DOCUMENT NODE
  echo '<marker ';
  echo 'name="' . parseToXML('&','&amp;', $row['name']) . '" ';
  echo 'address="' . parseToXML($row['address']) . '" ';
  echo 'lat="' . $row['lat'] . '" ';
  echo 'lng="' . $row['lng'] . '" ';
  echo 'type="' . $row['type'] . '" ';
  echo '/>';
}
// End XML file
echo '</markers>';
?>

Answer 1

尝试以下添加小更改的内容：

// Select all the rows in the markers table
$query = "SELECT * FROM markers";
$results = mysql_query($query);
if (!$results) {
  die('Invalid query: ' . mysql_error());
}
header("Content-type: text/xml");
// Start XML file, echo parent node
echo '<?xml version="1.0" encoding="UTF-8"?>' . "\r\n";
echo '<markers>' . "\r\n";
// Iterate through the rows, printing XML nodes for each
while ($row = mysql_fetch_assoc($results)){
    // ADD TO XML DOCUMENT NODE
    echo "\t<marker ";
    echo 'name="' . parseToXML('&','&amp;', $row['name']) . '" ';
    echo 'address="' . parseToXML($row['address']) . '" ';
    echo 'lat="' . $row['lat'] . '" ';
    echo 'lng="' . $row['lng'] . '" ';
    echo 'type="' . $row['type'] . '" ';
    echo "/>\r\n";
}
// End XML file
echo "</markers>\r\n";

这将添加适当的行格式和缩进，这适用于XML格式。 JS不会阅读它们，但这是一个很好的做法。然后，您还可以直接在浏览器中调用页面来测试结果，如果XML正确，则需要对JS代码进行故障排除，并了解它为什么不生成标记。检查您的浏览器控制台（或使用FireBug等工具）查找错误。

修改

发现<div id="map" style="width: 100%; height: 100%"></div>似乎是问题的一部分。建议更改为像素值：<div id="map" style="width: 1024px; height: 680px;"></div>。

您也未在load()中设置或更新标记数组：

map.markers = map.markers || [];

再次在downloadUrl()：

map.markers.push(marker);

工作示例：http://jsfiddle.net/Twisty/k5nnsqcL/

如何使用数据库和PHP在googlemap上创建标记

1 个答案: