未捕获的TypeError：无法读取未定义的属性

Question

我使用ajax创建了一个动态内容：

$(function(){
   $('#loadFeed').bind('click',function(){ 
        $.getJSON('getData.php', function(json) {
            var output="<ul id='feedsList'>";

            for(var i=json.posts.length-1;i>=json.posts.length-31;i--){
                output+="<li class='post'>";
                output+="<div class='text' id='"+json.posts[i].id+"'>"+json.posts[i].shortmsg+"</div>";         
                output+="</li>";
            }
            output+="</ul>"
            $(output).appendTo('.posts');
    });
  });
});

html代码：

<div class="posts">
    <!--dynamic content here-->
</div>

我想点击短文本，然后文本将展开以显示整个文本。实现它的代码是：

<script>
$(".posts").on("click", ".text",function(){
        var thisID = this.id; 
        $.getJSON('getData.php', function(json) {  
            alert(thisID);
            $(this).replaceWith("<div class='long_text'>"+json.posts[thisID-1].msg+"<div>"); 
        });
});
</script>

错误是

Uncaught TypeError: Cannot read property 'msg' of undefined .

我不知道为什么它将msg识别为未定义。我在getData.php中定义了msg，就像我对id，shortmsg所做的一样。这是我的getData.php：

<?php

        // Connecting, selecting database
        $link = mysql_connect('localhost', 'root')
            or die('Could not connect: ' . mysql_error());
        mysql_select_db('my_db') or die('Could not select database');

        // Get Data
        $query = 'SELECT * FROM feed ORDER BY feed.userID ASC';

        $result = mysql_query($query) or die('Query failed: ' . mysql_error());

        //create json format text from posts
        $json=array();
        while($row=mysql_fetch_array($result)){
            $post=array('id'=>$row['userID'],'avatar'=>$row['avatar'],'postType'=>$row['postType'],'msg'=>$row['msg'],'shortmsg'=>$row['shortmsg'],'title'=>$row['title'],'type'=>$row['type'],'pic'=>$row['pic'],'link'=>$row['link']);
            array_push($json,$post);
        }

        //display it to the user
        header('Content-type: application/json');
        echo "{\"posts\":".json_encode($json)."}";

        // Closing connection
        mysql_close($link);

        return $row;

?>

谢谢！