如何使用json encode和ajax返回html中的数据

Question

我正在尝试使用JSON和ajax显示来自我的数据库的数据，但我在执行该操作方面遇到了困难。请帮助，谢谢。

//我的js函数

function show_messages(){
     $.ajax({
         url: "display_table.php",
         cache: false,
         success: function(html){
             $("#table_content").html(html);
         }
     });
 }


 <?php    

$ query =“SELECT * FROM comments ORDER BY id ASC”; $ result = mysql_query（$ query）;

if(isset($_REQUEST['AnswerId'])){ $AnswerId = $_REQUEST['AnswerId']; }
else {$AnswerId = 0; }

$i=0;
while ($mytablerow = mysql_fetch_row($result)){
$mytable[$i] = $mytablerow; 
$i++;   
}

function tree($treeArray, $level, $pid = 0) {
global $AnswerId;
if (! $treeArray) 
{ return; } 
     foreach($treeArray as $item){
    if ($item[1] == $pid){
        ?>  
        <div class="CommentWithReplyDiv" style="margin-left:<?php echo($level*60);?>px">    
        <div class="CommentDiv">
        <div class="User"><?php echo($item[2]) ; ?></div>
        <div class="Message"><?php echo($item[3]) ; ?></div>
        <div class="Date"><?php echo($item[4]) ; ?></div>
        <?php               
if ($level<=10) 
{ echo '<a href="" class="ReplyLink" onclick="AnswerComment('.$item[0].');return false;">Reply</a>'; }
echo '<a href="" class="DeleteLink" onclick="DeleteComment('.$item[0].');return false;">Delete</a>';
        ?> </div> <?php

        if ($AnswerId == $item[0]){
            ?><div id="InnerDiv"><?php
            ShowForm($AnswerId);
            ?></div><?php   
        } 

        ?> </div> <?php 
        echo ('<br/>');
        tree($treeArray, $level+1, $item[0]);
    }       
} }
tree($mytable, 0);
?>