我試着在我的本地機器上運行Windows 7家庭高級版,Apache,PHP和MySQL,但它不會返回任何結果到瀏覽器的這個AJAX示例。在這裏讀了幾篇文章後,我下載螢火,從螢火蟲控制檯 - >所有我所得到的是這樣的:爲什麼不會這個AJAX腳本工作本地主機上運行
GET http://localhost/dev/ajax/ajax-example.php?age=100&wpm=100&sex=m 200 OK 1.01s
響應
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>
上面這個東西應該怎麼去broswer。該腳本似乎工作得很好,除了div不會刷新新的內容。
這是一個瀏覽器問題或Windows/JavaScript問題。我需要做些什麼來完成這項工作?能否請你幫忙。
這裏是教程頁,我從這裏得到了所有。
http://www.tutorialspoint.com/ajax/ajax_database.htm
本頁面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>
這個網頁是Ajax的使用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;
?>
如果它是微不足道的,爲什麼使用jQuery?有點像打開胡桃木很容易,所以使用大錘讓它更容易。 – epascarello 2012-07-19 12:11:36
您的建議工作。感謝您的幫助。它實際上是該網站的錯誤。它的ajaxDisplay.value = ajaxRequest.responseText;在這個例子中,但是ajaxDisplay.innerHTML = ajaxRequest.responseText;在他們的網站實際使用的代碼中。 – Norman 2012-07-19 12:17:35
2 epascarello 這是微不足道的,但對於作者不是,用jquery他可以做到這一點3次少代碼,對於最終用戶它是沒有問題,等待更多0.01s – 2012-07-19 12:22:44