使用sql/google地图教程中的phpsqlajax.php脚本,似乎我已经设法用mysql的值生成xml元素(标记)。尽管没有内容被回显到页面,但标记列表正在根据控制台日志通过 mysql 填充。
1. 这是否意味着 xml 正在正确生成?
我问是因为phpsqlajax_map_v3.php脚本收到一个错误报告:计算 XML 文档元素给出类型问题 null 不是对象。
2. 这似乎是在说标记节点是空的,因此没有从 mysql 获取值。
代码:
\\\已编辑的工作脚本\\\
phpsqlajax.php Google 教程链接
<?php
// Start XML file, create parent node
$dom = new DOMDocument("1.0");
$node = $dom->createElement("markers");
$parnode = $dom->appendChild($node);
?>
<?
// connect
$host = "localhost";
$username = 'root';
$psswrd = 'root';
$db = 'sql_maps';
$dbc = mysqli_connect($host, $username, $psswrd, $db);
if(!mysqli_connect_errno() ) {
} else {
die("Database failed: " .
mysqli_connect_error() .
" ( " . mysqli_connect_errno() . " )"
);
}
?>
<?php
// Using PHP's domxml Functions to Output XML
// Select all the rows in the markers table
// get data
$query = "SELECT * FROM markers WHERE 1";
// catch resource(collection of database rows)
$result = mysqli_query($dbc, $query);
// check
if($result) {
} else {
die("connection failed");
}
header("Content-type: application/xml");
while ($row = mysqli_fetch_assoc($result)){
// ADD TO XML DOCUMENT NODE
$node = $dom->createElement("marker");
$newnode = $parnode->appendChild($node);
$newnode->setAttribute("name",$row['name']);
$newnode->setAttribute("address", $row['address']);
$newnode->setAttribute("lat", $row['lat']);
$newnode->setAttribute("lng", $row['lng']);
$newnode->setAttribute("type", $row['type']);
}
echo $dom->saveXML();
?>
phpsqlajax_map_v3.php
<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 Example</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript"
src="https://maps.googleapis.com/maps/api/js?key=keyIntentionallyLeftOut&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'
}
};
var map;
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("../geo_scripts/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>
3.由于这是从脚本生成xml,我将如何将浏览器指向xml以查看它是否有效或有效?
我已经在stackoverflow上搜索了类似的问题,但是似乎这些修复程序在这里不适用,即内容类型:xml,谷歌代码教程中的各种语法错误,代码结构等。
删除echo 'sueccess';
并echo 'success';
。
XML文档必须以<?xml
开头,当您使用这些行时,它将以sueccesssuccess<?xml
开头,解析失败。
当您不想收到有效的 XML 文档时,不要在页面内回显任何内容($dom->saveXML()
除外)。