在本地使用 xml 文件(从 php 生成)时，此代码似乎工作得很好。即/phpsqlajax_genxml.php
但是，当我尝试在外部使用相同的 xml 文件时，它不会填充。即http://brewslocal.com/phpsqlajax_genxml.php
有什么帮助可以解释为什么吗？下面附上我正在使用的代码。
<!DOCTYPE html>
<head>
<link rel="icon" href="beer-128.png" >
<title>BrewsLocal - Find More Local Breweries</title>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?libraries=places&sensor=false"></script>
<script type="text/javascript">
//<![CDATA[
var customIcons = {
Brewery: {
icon: '/Logomakr_5xi5EC.png'
},
Brewpub: {
icon: '/Logomakr_5xi5EC.png'
}
};
var gmarkers1 = [];
function getParameterByName(name, url) {
if (!url) url = window.location.href;
name = name.replace(/[\[\]]/g, "\\$&");
var regex = new RegExp("[?&]" + name + "(=([^&#]*)|&|#|$)"),
results = regex.exec(url);
if (!results) return null;
if (!results[2]) return '';
return decodeURIComponent(results[2].replace(/\+/g, " "));
}
var currentlat = getParameterByName('lat');
var currentlng = getParameterByName('lng');
function initialize() {
var map = new google.maps.Map(document.getElementById("map"), {
center: new google.maps.LatLng(currentlat , currentlng),
zoom: 13,
mapTypeId: 'roadmap'
});
var infoWindow = new google.maps.InfoWindow;
// Change this depending on the name of your PHP file
downloadUrl("/phpsqlajax_genxml.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 id = markers[i].getAttribute("id");
var address = markers[i].getAttribute("longAddress");
var lat = markers[i].getAttribute("lat");
var lng = markers[i].getAttribute("lng");
var type = markers[i].getAttribute("type");
var logo = markers[i].getAttribute("logo");
var point = new google.maps.LatLng(
parseFloat(markers[i].getAttribute("lat")),
parseFloat(markers[i].getAttribute("lng")));
var html = "<div class='infoblock'><img style='display: block; margin: auto;'src='" + logo + "' height='120px' width='auto'/><h2>" + name + "</h2><a href='/brewery?id=" +
id + "'>Brewery Page</a><a href='http://www.google.com/maps/?q=" + point
+ "' target='_blank' style='opacity: .8;'>Location</a></div>";
var icon = customIcons[type] || {};
var marker = new google.maps.Marker({
map: map,
position: point,
icon: icon.icon,
type: type,
origin: new google.maps.Point(0, 0),
anchor: new google.maps.Point(0, 0),
shadow: icon.shadow
});
gmarkers1.push(markers);
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('phpsqlajax_genxml.php') :
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="initialize()">
<?php include("header.php"); ?>
<div id="map"></div>
<div style=" position: fixed;
bottom: 0;
width: 100%;
z-index: 9;
margin: 0;
padding: 0;" ><h2 id="error"></h2><input type="text" name="location" id="location" class="main-input" placeholder="Enter A Location" required><button type="submit" onclick="testlocation()" class="btn" id="location-button" >Locate</button></div>
</div>
</body>
</html>最佳答案
您正在访问 Ajax Cross Origin。
https://en.wikipedia.org/wiki/Cross-origin_resource_sharing
您可以在提供 phpsqlajax_genxml.php 的服务器中使用 Access-Control-Allow-Origin header ，尝试添加
header('Access-Control-Allow-Origin: *');
在 phpsqlajax_genxml.php 的开头也许会有帮助。
或者您可以使用 JSONP 的替代方法(但这需要针对您的情况进行更多开发。)
关于php - MySQL 和 PHP 与 Google map - 数据源仅在本地工作，无法从 http url 访问时工作，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/41724330/