【发布时间】：2012-07-19 11:54:51
【问题描述】：
我在运行 Windows 7 家庭高级版、apache、php 和 mysql 的本地计算机上尝试了这个 ajax 示例，但它不会向浏览器返回任何结果。在这里阅读了几篇文章后，我下载了firebug，并从firebug控制台->我得到的都是这个：
GET http://localhost/dev/ajax/ajax-example.php?age=100&wpm=100&sex=m 200 OK 1.01s
回应
Query: SELECT * FROM ajax_example WHERE sex = 'm' AND age <= 100 AND wpm <= 100
<br /><table><tr><th>Name</th><th>Age</th><th>Sex</th><th>WPM</th>
</tr><tr>td>Frank</td><td>45</td><td>m</td><td>87</td></tr><tr><td>Regis</td>
<td>75</td><td>m</td><td>44</td></tr></table>
上面的这个东西应该是去浏览器的。该脚本似乎工作正常，除了 div 不会刷新新内容。
这是浏览器问题还是 windows/javascript 问题。我需要做什么才能使它正常工作？你能帮忙吗？
这是我从中获得所有这些的教程页面。
http://www.tutorialspoint.com/ajax/ajax_database.htm
这个页面是ajax.html
<html>
<body>
<script language="javascript" type="text/javascript">
<!--
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.value = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var age = document.getElementById('age').value;
var wpm = document.getElementById('wpm').value;
var sex = document.getElementById('sex').value;
var queryString = "?age=" + age ;
queryString += "&wpm=" + wpm + "&sex=" + sex;
ajaxRequest.open("GET", "ajax-example.php" + queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
<form name="myForm">
Max Age: <input type="text" id="age" /> <br />
Max WPM: <input type="text" id="wpm" />
<br />
Sex: <select id="sex">
<option value="m">m</option>
<option value="f">f</option>
</select>
<input type="button" onclick="ajaxFunction()" value="Query MySQL"/>
</form>
<div id="ajaxDiv">Your result will display here</div>
</body>
</html>
这个页面是ajax-example.php
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "norman";
$dbname = "test";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$age = $_GET['age'];
$sex = $_GET['sex'];
$wpm = $_GET['wpm'];
// Escape User Input to help prevent SQL Injection
$age = mysql_real_escape_string($age);
$sex = mysql_real_escape_string($sex);
$wpm = mysql_real_escape_string($wpm);
//build query
$query = "SELECT * FROM ajax_example WHERE sex = '$sex'";
if(is_numeric($age))
$query .= " AND age <= $age";
if(is_numeric($wpm))
$query .= " AND wpm <= $wpm";
//Execute query
$qry_result = mysql_query($query) or die(mysql_error());
//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Age</th>";
$display_string .= "<th>Sex</th>";
$display_string .= "<th>WPM</th>";
$display_string .= "</tr>";
// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$display_string .= "<tr>";
$display_string .= "<td>$row[name]</td>";
$display_string .= "<td>$row[age]</td>";
$display_string .= "<td>$row[sex]</td>";
$display_string .= "<td>$row[wpm]</td>";
$display_string .= "</tr>";
}
echo "Query: " . $query . "<br />";
$display_string .= "</table>";
echo $display_string;
?>
【问题讨论】：
标签： php javascript mysql ajax apache