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我很難得到這個答案...我可以用xmlhttprequest獲取什麼？
我能夠從一個PHP文件與Ajax的HTML元素？我有一種感覺，我試圖做一些我不應該做的事情。
我使用：
.
.
.
<tr>
<td colspan="3" style="font-size:11px;"><a href="javascript:void(0)" onclick="getData('includes/forgot-passwordajax.php', 'targetDiv')">Forgot Password?</a></td>
</tr>
</table></td>
</tr>
</table>
</form>
<div id="targetDiv">
</div>
，包括/忘記-passwordajax.php與表單數據的表格，等難道這工作？
編輯：這是腳本：
<script language = "javascript">
var XMLHttpRequestObject = false;
if (window.XMLHttpRequest) {
XMLHttpRequestObject = new XMLHttpRequest();
} else if (window.ActiveXObject) {
XMLHttpRequestObject = new ActiveXObject("Microsoft.XMLHTTP");
}
function getData(dataSource, divID)
{
if(XMLHttpRequestObject) {
var obj = document.getElementById(divID);
XMLHttpRequestObject.open("GET", dataSource);
XMLHttpRequestObject.onreadystatechange = function()
{
if (XMLHttpRequestObject.readyState == 4 &&
XMLHttpRequestObject.status == 200) {
obj.innerHTML = XMLHttpRequestObject.responseText;
}
}
XMLHttpRequestObject.send(null);
}
}
</script>
PHP：
<?php
<div class='a_login'>
<table class="tabbody">
<tr class="tprowbgcolor">
<td class="title">Retrieve Password</td></tr>
</table>
<center><BR><BR>
<table cellpadding="0" cellspacing="0">
<form name='loginForm' action="forgot-password.php" method="POST">
<tr><td><img src="<?=SITE_URL?>/images/top-bg.jpg"></td></tr>
<tr>
<td class="a_td_color" align="center" valign="top">
if(isset($_POST['formSubmitted']))
{
$email = addslashes(trim($_POST['e_address']));
if($email=="") {
echo '<p>Just kidding? Ehh! <a href="forgot-password.php">try again</a></p>';
} else {
$query = "SELECT `Password`, `UserName` FROM users WHERE `Email`='$email'";
$result = mysql_query($query);
if(mysql_num_rows($result)>0) {
$row = mysql_fetch_assoc($result);
$password=$row['Password'];
$username=$row['UserName'];
$to = $email;
$headers = 'MIME-Version: 1.0' . "\r\n";
$headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n";
$headers .= 'To: '.$username.' <'.$email.'>' . "\r\n";
$headers .= 'From: Joel Laviolette <[email protected]>' . "\r\n" .
'X-Mailer: PHP/' . phpversion();
$subject = "Here is your password";
$message = '<p>Hello '.$username.',</p><p>Your password is: ';
$message .= '<strong>'.$password.'</strong><br><br>';
$message .= 'You can login here: <a href="'.SITE_URL.'/login2.php">'.SITE_URL.'/login2.php</a></p>';
$message .= '<p>With Love and gratitude,<br>Joel Laviolette</p>';
if(mail($to, $subject, $message, $headers)) {
echo '<p>Your password has been sent. Please check your email.</p>';
} else {
echo '<p>Email could not be sent! Please contact me directly at [email protected]</p>';
}
} else {
echo '<p>Email not found! <a class="external" href="forgot-password.php">try again</a></p>';
}
}
} else {
<table border='0'>
<tr>
<td colspan="3">
<p>Email Address<br />
<input type="text" name="e_address" value="" size="20"></p></td>
</tr>
<tr>
<td colspan="3" align="center"><input type="submit" value="Submit" name="submit"><input type="hidden" name="formSubmitted" value='1'></td>
</tr>
<tr>
<td colspan="3"> </td>
</tr>
</table>
?>
您應該能夠獲取數據並根據它更改HTML - 這就是AJAX的用途。但是你可以發佈'getData（）'的定義嗎？因爲它不清楚在這裏做了什麼？ – 2010-01-27 06:01:45
完成。我仍然不相信我的php文件有任何形式的正確格式，但我不願意發佈它 - 哈哈 - 我沒有寫它。 – Joel 2010-01-27 06:04:43
您發佈的所有代碼_looks_正確，發佈PHP。可能有個問題， – 2010-01-27 06:08:31