简单的JOSN实现

<?php 
if($_GET["opp"]=="getjosn") 
{ 
    //$a=array("title"=>"jjss"); 
    echo '{ "questionid":"1178", 
            "title":"quick speach888", 
            "a":"1995yeahs", 
            "b":"ten words", 
            "c":"", 
            "d":"", 
            "userid":"0", 
            "right":"1", 
            "username":"", 
            "categoryid":"1", 
            "color":"red", 
            "tsjifen":"0", 
            "catelog_id":"29", 
            "is_pass":"1", 
            "flag":"0"}'; 
    //echo json_encode($a); 
    exit; 
} 
?> 
<script language=javascript src="jquery.js"></script> 
<SCRIPT LANGUAGE="JavaScript"> 
<!-- 
$(document).ready(function() { 
    $.ajax({ 
        type: "get",//使用get方法访问后台 
        dataType: "json",//返回json格式的数据 
        url: "json.php?opp=getjosn",//要访问的后台地址 
        success: function(msg){ 
        alert(msg.title); 
        } 
    }) 
}) 
//--> 
</SCRIPT>