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我一直有問題，試圖提交一個使用jquery ajax，php到mysql數據庫的表單。我重複使用了一個我在網上找到的動畫形式的代碼。 HTML代碼如下所示（文件名： 「Slider.html」）：Submtting使用jquery ajax和php的mysql數據庫的動畫形式
<html>
<body>
<div id="form_wrapper" class="form_wrapper">
<form class="register" style="right:-500px;backgroundcolor:grey;">
<h3>Register</h3>
<div class="column">
<div>
<label>First Name:</label>
<input type="text" />
<span class="error">This is an error</span>
</div>
<div>
<label>Last Name:</label>
<input type="text" />
<span class="error">This is an error</span>
</div>
</div>
<div class="column">
<div>
<label>Username:</label>
<input type="text"/>
<span class="error">This is an error</span>
</div>
<div>
<label>Email:</label>
<input type="text" />
<span class="error">This is an error</span>
</div>
<div>
<label>Password:</label>
<input type="password" id="r_passwordField"/>
<span class="error">This is an error</span>
</div>
</div>
<div class="bottom">
<div class="remember">
<input type="checkbox" />
<span>Send me updates</span>
</div>
<input type="submit" id="registrationform" value="register"/>
<a href="index.html" rel="login" class="linkform">You have an account already? Log in
here</a>
<div class="clear"></div>
</div>
</form>
<form class="login active">
<h3>Login</h3>
<div>
<label>Username:</label>
<input type="text" />
<span class="error">This is an error</span>
</div>
<div>
<label>Password: <a href="forgot_password.html" rel="forgot_password" class="forgot
linkform">Forgot your password?</a></label>
<input type="password" id="l_passwordField" size="10"/>
<span class="error">This is an error </span>
</div>
<div class="bottom">
<div class="remember"><input type="checkbox" /><span>Keep me logged in</span></div>
<input type="submit" id="loginform" value="Login"></input>
<a href="register.html" rel="register" class="linkform">You don't have an account yet?
Register here</a>
<div class="clear"></div>
</div>
</form> <form class="forgot_password">
<h3>Forgot Password</h3>
<div>
<label>Username or Email:</label>
<input type="text" />
<span class="error">This is an error</span>
</div>
<div class="bottom">
<input type="submit" id="forgortform" value="Send reminder"></input>
<a href="index.html" rel="login" class="linkform">Suddenly remebered? Log in here</a>
<a href="register.html" rel="register" class="linkform">You don't have an account?
Register here</a>
<div class="clear"></div>
</div>
</form>
</div>
<div id="graph-wrapper" class="graph-container" style="display:none">
<h3>Standard Diviation Bell Curve<h3>
<div class="graph-info">
<a href="javascript:void(0)" class="visitors">Visitor</a>
<a href="#" id="lines" class="active"><span></span></a>
</div>
<div class="graph-container">
<div id="graph-lines"></div>
<div id="graph-bars"></div>
</div>
</div>
</div>
</body>
</html>
jQuery的Ajax代碼如下（在同一個文件的HTML代碼jQuery腳本）：
<script>
$(document).ready(function(){
//animated code and other codes omitted
("#registrationform").click(function(){
$.post("postdata.php",
{fname:"Ibrahim",lname:"Ahmadu",email:"[email protected]",uid:"ibro2stay",pwd:"ibro2stay",mean:"0.1",varience:"0.1",sdev:"0.
1",duration:"0.1"},function(responce){
if(responce==0){
alert("There was an error updating the record");
}else{
alert("update successful");
}
});
});
});
</script>
下面是PHP代碼（PHP代碼的文件名： 「postdata.php」）：
<?php
$fname=$_REQUEST["fname"];
$lname=$_REQUEST["lname"];
$email=$_REQUEST["email"];
$uid=$_REQUEST["uid"];
$pwd=$_REQUEST["pwd"];
$mean=$_REQUEST["mean"];
$varience=$_REQUEST["varience"];
$sdev=$_REQUEST["sdev"];
$duration=$_REQUEST["duration"];
$con=mysql_connect('localhost','root','');
if(!$con)
{
die("Error Connecting to database;"+mysql_error());
}
$database=mysql_select_db('mydb');
if(!$database)
{
die("Error Connecting to database;"+mysql_error());
}
$update = mysql_query("insert into users values('$fname','$lname','$email','$uid','$pwd','$mean','$varience','$sdev','$duration')");
if(!$update)
{
die("Update wasn't Success full"+mysql_error());
}
echo "update successfull";
mysql_close($con);
?>
每當我點擊註冊按鈕沒有任何反應。該頁面僅刷新回登錄表單，因爲它具有默認表單的「活動」類，並且瀏覽器地址欄從該URL：「localhost/slider.html」更改爲以下URL：「localhost/slider.html？」 。
我希望我的問題足夠明確，因爲我需要一個緊急答案，因爲這是我的論文項目，而且我的選擇不多。
非常感謝您的及時響應，無論如何我早就明白這一點，但感謝提示.....這聽起來可能很有趣，但我直接在記事本上編碼....你有鏈接到任何可靠的在線js代碼debuger？ – user28864
你不需要一個。只需檢查默認包含在任何現代brwoser中的javacript控制檯，或作爲擴展。 Chrome已內置，FireFox有螢火蟲，歌劇有DragonFly ....等 – itachi
感謝您的頭....乾杯 – user28864