返回数据会跳过jquery ajax函数的div

Question

$(document).ready(function()
{
var ad_id =/*$("#ad_id").val() */<?php echo $ad['id']; ?>;
var uid = /*$("#ad_id_poster").val()*/ <?php echo $ad['uid']; ?>;

$.ajax(
{
    type:'POST',
    url:'../business_logic/answer.php',       //URL
    data:{  id: ad_id, ad_uid:uid},  //Data
    beforeSend:function(){
        $('#wait').show();
                },
    success:function(data,status){                      //Data,status
    var item = $(data).hide().slideDown("slow");
    $('#wait').hide();
    $('.all_qa_updates').prepend(item);

    // reset form and button
    $('#cmnt_in').val('');                       
    }
});
});

此代码从answer.php文件返回数据

 $n = null;
     $id =  $_POST['id'] ; //Ad Id
     $ad_uid = $_POST['ad_uid']; //Ad poster ID 
     $QueAns = new QueAns($n,$n,$n,$n,$n,$n,$n);
     $all_aq = $QueAns->get_ques($id); //param ad id
     $current_u = $_SESSION['uid'];

if(isset($all_aq) === true)
{
    for( $i = 0 ; $i < count($all_aq) ; $i++ )
    {
        $name = $all_aq[$i]['u_fn'].' '.$all_aq[$i]['u_ln'];
        $q = $all_aq[$i]['q_ques'];
        $path = $dp_path.$all_aq[$i]['u_img'];
        $check_ans = $all_aq[$i]['ans']['fn'];

        ?>
            <div class="all_qa panel panel-default">
                <img src="<?php 
                    if(isset($all_aq[$i]['u_img'])){
                        echo $path;
                    }
                    else{
                        echo '../images/default_profile_pic/d_boy.png';
                    }
                ?>">
                <a href="#"><?php echo $name;?></a><br>
                <p><?php echo $q;?></p>

        <?php

            /* if((isset($all_aq[$i]['ans']['fn']) === true) && (isset($_SESSION['uid']) ===          true))
            {
                echo '</div>';
            } */
            if(isset($check_ans)) 
            {?>
                </div>
                <?php //echo 'ans';
                $a_name = $all_aq[$i]['ans']['fn'].' '.$all_aq[$i]['ans']['ln'];
                $a_ans = $all_aq[$i]['ans']['ans'];
                $a_path =  $dp_path.$all_aq[$i]['ans']['u_img'];
                echo '<div class="ans panel panel-default">';
                    echo '<img src="'.$a_path.'">';
                    echo "<a href='#'>".$a_name."</a><br>";
                    echo '<p>'.$a_ans.'</p>';
                echo '</div>';
            }
            if(isset($check_ans) === false && 
                $ad_uid === $current_u)
            {
                echo '<p>reply</p>';
                echo '<input value="Reply" type="button" data-toggle=btn_'.$all_aq[$i]['q_id'].' id="reply">';
                ?>
                </div>
                <form class="answer_form" id="btn_<?php echo $all_aq[$i]['q_id'];?>">
                    <img src='../images/default_profile_pic/d_boy.png'>
                    <textarea id="cmnt_in" type="text" name="comment" placeholder="Ask Question..." required></textarea>
                    <input value="POST" type="button" id="send_data" onclick="send_data" class="btn p_srch">
                    <input type="hidden" id="q_id" value="<?php echo $all_aq[$i]['q_id'];?>" />
                </form>
        <?php       
            }

    }
}

问题是，当jquery返回数据时，它会跳过第一个div以及它的结束div ....请帮助。

另一个奇怪的事情是，当我明确地为变量id和ad_id赋值时，返回数据完全按照我想要的方式显示，并且没有丢失任何div，但是当它从帖子中获取时跳过一些div.pls帮助