关于.Ajax的使用

都说使用jquery可以简化Ajax，我就用的，但是搞了一天都是出错，晕死了，请大家帮忙看一下


前台很简单，代码如下

先引入js文件
<script language="javascript" src="jquery .js"></script>
<script language="javascript" src="check.js"></script>


                <form id="form1" name="form1" method="post">
                    Username: <br />
                    <input type="text" id="username" name="username" /><br /><br />
                     Password: <br />
                    <input type="password" id="password" name="password" /><br /><br />
                    <button type="submit" id="submit" name="submit"><img src="images/open.gif" /> Login</button>
                    <button type="reset" id="reset" name="reset"><img src="images/open.gif" /> Reset</button><br /><br />
               </form>
              <div class="mid_li_2"></div>

jquery .js就不用说了，是原版的，不用动。那么我就继续下了check.js文件，代码如下

    $("#submit").click(function(){
        if($("#username").val()==""){
            $(".mid_li_2")
                .slideDown("slow")
                .html("<span style='font-size:12px; color:#333; text-align:center'><center>Username can't be empty !</center></span>")
                return false;
        }else if($("#password").val()==""){
            $(".mid_li_2")
                .slideDown("slow")
                .html("<span style='font-size:12px; color:#333; text-align:center'><center>Password can't be empty !</center></span>")
                return false;
        }else{
            var dataString = 'username='+ $("#username").val() + '&password=' + $("#password").val();
            $.ajax({
                type: "POST",
                url: "user.php",
                data: dataString,
                error: function(){alert('Error loading XML document');},
                success: function(data) {
                    alert(data);
                    if(data == 1){
                        alert("successful");
                    }else if(data == 0){
                        $(".mid_li_2")
                            .slideDown("slow")
                            .html("<span style='font-size:12px; color:#333; text-align:center'><center>No match for Username and / or Password.</center></span>")
                            return false;
                    }
               }
             });
        }
    })

这里用了Ajax，user.php文件 代码如下

<?php
$username = $_POST['username'];
$password = $_POST['password'];
if($username == "gideon300"){
    echo "1";
}else{
    echo "0";    
}
?>

其实很简单，但是我就是不知道错在哪里，总是不能成功，谁可以提示一下啊？

type不要用submit,你用button试度

果然是！  谢谢，太感谢了！原来如此啊~       

复制代码 //也可以使用submit,代码如下： $("#myform").submit(function(){    .............   return false; });

谢谢你，学习到很多！