Ajax函数第一次工作，但第二次没有？

Question

这是我的index.html文件

<form method="post" id="form" action="join.php">
<ul>
    <li><input id="name" name="name" type="text" /></li>
    <li><input id="username" name="username" type="text" /></li>
    <li><input id="password" name="password" type="password" /></li>
    <li>
        <select id="gender" name="gender"> 
            <option value="">Gender</option>
            <option value="1">Male</option>
            <option value="2">Female</option>
        </select>
    </li>
</ul>
<div>
    <input type="submit" value="Submit" class="submit"/>
    <span class="error" style="display:none"> Please Enter Valid Data</span>
    <span class="success" style="display:none"> Registration Successfully</span>
</div>
</form>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
</script>
<script type="text/javascript" >
$(function() {
    $(".submit").click(function() {
        var name = $("#name").val();
        var username = $("#username").val();
        var password = $("#password").val();
        var gender = $("#gender").val();
        var dataString = 'name='+ name + '&username=' + username + '&password=' + password + '&gender=' + gender;

        if(name=='' || username=='' || password=='' || gender=='') {
            $('.success').fadeOut(200).hide();
            $('.error').fadeOut(200).show();
        } else {
            $.ajax({
                type: "POST",
                url: "join.php",
                data: dataString,
                success: function(){
                $('.success').fadeIn(200).show();
                $('.error').fadeOut(200).hide();
                }
            });
        }
        return false;
    });
});
</script>

这是join.php

<?php
$conn = mysql_connect('localhost','root','') or die (mysql_error);
$db=mysql_select_db('test', $conn) or die (mysql_error);
if($_POST) {
    echo"processing...........";
    $name=$_POST['name'];
    $username=$_POST['username'];
    $password=$_POST['password'];
    $gender=$_POST['gender'];
    mysql_query("INSERT INTO user1 (name,username,password) VALUES('$name','$username','$password')");
} else {
    echo"unsuccessful";
}
?>

我想将名称，用户名和密码的值存储在数据库中，但它仅在第一次尝试时存储。当我第二次提交时没有任何反应。那是为什么？