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我的 ajax 成功函数在将数据发送到我的数据库后不执行脚本，数据在我的数据库中保存得很好，但之后一切都停在那里。可能有什么问题？

Javascript代码：

$(document).ready(function() {
    $(".wall_update").click(function() {
        var element = $(this);
        var boxval = $("#content").val();
        var dataString = 'content=' + boxval;


        $("#form1").validationEngine({
            type: "POST",
            ajaxSubmit: true,
            ajaxSubmitFile: "update_ajax.php",
            cache: false,
            success: function(html) {
                alert('success');
                $("ol#update").prepend(html);
                $("ol#update li:first").slideDown("slow");
                document.getElementById('content').value = '';
                $('#content').value = '';
                $('#content').focus();
                $("#flash").hide();

            }
        });

    });
});​

update_ajax.php

    <?php
    include("db.php");
    include("tolink.php");


    if(isSet($_POST['content']))

    {
    $id=time();//Demo Use

    $msg=$_POST['content']; 
    $date=date("M j, Y ");
    $sql=mysql_query("insert into appointments(message,date_sent)values('$msg','$date')");
    $result=mysql_query("select * from appointments order by msg_id desc");
    $row=mysql_fetch_array($result);
    $id=$row['msg_id'];
    $msg=$row['message'];
    $date=$row['date_sent'];

    $msg= nl2br($msg);
    $msg="<br>{$msg}<br>{$date}";
    }


    ?>




    <li class="bar<?php echo $id; ?>">
    <div align="left" class="post_box">
    <span style="padding:10px"><?php echo $msg; ?> </span>
    <span class="delete_button"><a href="#" id="<?php echo $id; ?>" class="delete_update">X</a></span>
    <span class='feed_link'><a href="#" class="comment" id="<?php echo $id; ?>">comment</a></span>
    </div>
    <div id='expand_box'>
    <div id='expand_url'></div>
    </div>
    <div id="fullbox" class="fullbox<?php echo $id; ?>">
    <div id="commentload<?php echo $id; ?>" >

    </div>
    <div class="comment_box" id="c<?php echo $id; ?>">
    <form method="post" action="" name="<?php echo $id; ?>">
    <textarea class="text_area" name="comment_value" id="textarea<?php echo $id; ?>">
    </textarea><br />

<input type="submit" value=" Comment " class="comment_submit" id="<?php echo $id; ?>"/>
</form>
</div>
</div>


</li>

HTML 脚本

<div align="left">
<form  method="post" name="form" action="" id="form1">
<table cellpadding="0" cellspacing="0" width="500px">

<tr><td align="left"><div align="left"><h3>What are you doing?</h3></div></td></tr>
<tr>
<td style="padding:4px; padding-left:10px;" class="update_box">
<textarea  class="validate[custom[last]] text-input" cols="30" rows="10" style="width:480px;font-size:14px; font-weight:bold" name="content" id="content"  ></textarea><br />
<input type="submit"  value="Update"  name="submit" class="wall_update"/>
</td>

</tr>

</table>
</form>

</div>

<ol  id="update" class="timeline">
// Display here after sending data to the database
</ol>
4

1 回答 1

2 
$('#content').value = '';

jQuery 没有value属性，你应该使用val：

$('#content').val('');

value是 DOM Input 对象的一个​​属性。

于 2012-11-15T16:45:25.520 回答