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我几乎复制了教程中的代码： https://developers.google.com/maps/articles/phpsqlajax_v3 （当然，我做了适当的修改，我猜）。谷歌地图API：空白页面（没有地图）
你能帮我弄清楚为什么地图不显示出来吗？
我使用的MySQL/PHP生成an XML output（类似代码的教程）：
<?php
require("database.php");
function parseToXML($htmlStr)
{
$xmlStr=str_replace('<','<',$htmlStr);
$xmlStr=str_replace('>','>',$xmlStr);
$xmlStr=str_replace('"','"',$xmlStr);
$xmlStr=str_replace("'",''',$xmlStr);
$xmlStr=str_replace("&",'&',$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 tbl_address table
$query = "SELECT * FROM tbl_address 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($result)){
// ADD TO XML DOCUMENT NODE
echo '<marker ';
echo 'name="' . parseToXML($row['address']) . '" ';
echo 'address="' . parseToXML($row['address']) . '" ';
echo 'lat="' . $row['latitude'] . '" ';
echo 'lng="' . $row['longitude'] . '" ';
echo 'type="USC Student Housing" ';
echo '/>';
}
// End XML file
echo '</markers>';
?>
这里是the map（类似代码的教程）：
<!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="http://maps.googleapis.com/maps/api/js?sensor=false"></script>
<script type="text/javascript">
//<![CDATA[
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(34.026485, -118.283794),
zoom: 14,
mapTypeId: 'roadmap'
});
var infoWindow = new google.maps.InfoWindow;
// Change this depending on the name of your PHP file
downloadUrl("http://firstchoicehousing.com/_api/google/maps/phpsqlajax_genxml2.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: 500px; height: 300px"></div>
</body>
</html>
感谢@ user1289347我是能够使地图的工作。最后一件事......我如何让这张地图看起来就像这样？：http://tiny.cc/yqg1cw（我需要右侧的地址列表和工具提示以包含网址等） – Omar 2012-04-20 00:53:56
注意！如果您正在阅读本文，我想评论一下谷歌的教程示例带有几个错误。大多数这些地址在这里解决 – Omar 2012-04-25 00:19:23
在谷歌的网站，示例代码[phpsqlajax_genxml.php]（http://gmaps-samples-v3.googlecode.com/svn/trunk/articles/phpsqlajax/phpsqlajax_genxml.php），以生成一个xml数据源文件有以下错误：'$ connection = mysql_connect（localhost，$ username，$ password）;'变量'localhost'应该是'$ localhost'（它缺少'$'-dollar标志）。另外，'$ localhost'需要包含在文件[phpsqlajax_dbinfo.php]中（http://gmaps-samples-v3.googlecode.com/svn/trunk/articles/phpsqlajax/phpsqlajax_dbinfo.php） – Omar 2012-04-25 00:30:52