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我在测试页面上测试了这段代码,但现在想在覆盖层(Lightview)调用的页面上实现它。地图不显示,我不知道为什么。覆盖图将空间大小映射到地图的大小,但没有显示。下面是代码:谷歌地图无法加载
<head>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />
<meta http-equiv="content-type" content="text/html; charset=UTF-8"/>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"> </script>
<?php
require_once('../maps/google.php');
if(isset($_POST['submit']))
{
$zipcode = $_REQUEST['zipcode'];
$lookupPerformed = false;
if (strlen($zipcode) > 0) {
$geocoder = new Geocoder('mykey');
try {
$placemarks = $geocoder->lookup($zipcode);
}
catch (Exception $ex) {
echo $ex->getMessage();
exit;
}
$lookupPerformed = true;
}
foreach ($placemarks as $placemark) {
$lat = $placemark->getPoint()->getLatitude();
$long = $placemark->getPoint()->getLongitude();
}
?>
<script type="text/javascript">
//<![CDATA[
var customIcons = {
A Company: {
icon: 'http://labs.google.com/ridefinder/images/mm_20_blue.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(<?php echo $lat;?>, <?php echo $long;?>),
zoom: 10,
mapTypeId: 'roadmap'
});
var infoWindow = new google.maps.InfoWindow;
// Change this depending on the name of your PHP file
downloadUrl("../testxml.php?zipcode=<?php echo $zipcode;?>", 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>
///// other PHP code
</head>
<form method="POST" id="ajaxForm" onsubmit="submitAjaxFormDemonstration()">
<input type="text" size="10" maxlength="10" name="zipcode" tabindex="1" value=" <?php echo $_POST['zipcode'];?>" />
<input type="submit" id="submit" value="Search" name="submit" tabindex="2" />
</form>
<body onload="load()">
<div id="map" style="width: 500px; height: 300px"></div>
</body>
</html>
感谢您的答案,但这并不能解决任何问题。 ../maps/google.php是地理编码器。我回合拉长,两者都在工作。我知道表单位于body标签之外,但对没有显示的地图没有影响。有人与谷歌地图或原型经验的答复? – savagenoob 2012-01-05 04:57:03
我确实有Google地图体验,谢谢。您是否能够显示简单/标准的地图,而不使用Geocoder?如果是这样,Geocoder(无论你从哪里得到)都是问题所在。如果您无法以任何方式查看地图,那么这是您的HTML问题。你不能在这里转储你的代码,只能显示它的一部分,并期望人们发现问题。 – 2012-01-05 05:15:17