jquery的ajax返回的如果是json格式，在success方法获取的值并不能直接当做json object处理，需要用$.parseJSON来转化成个json对象
例如
后台程序
<?PHP include ("json_helper.PHP"); $Josn = new Services_JSON (); $r = Array ("status"=>1,"list"=> array( Array ("c_id" => 1,"c_parent_id" => 0,"c_name" => "旅游美食","c_status" => 0,"c_type" => 0,"c_area" => "反选:福建|湖南|内蒙古|" ),Array ("c_id" => 2,Array ("c_id" => 3,Array ("c_id" => 4,) ); $r_josn = $Josn->encode($r); echo $r_josn;
静态页面
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <Meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>Insert title here</title> <script type="text/javascript" src="/js/jquery-1.4.2.min.js"></script> <script type="text/javascript"> $(function(){ $("#getjson").bind("click",function(){ $.ajax({ type: "POST",url: "/json/index.PHP",success: function(msg){ var r = $.parseJSON(msg); var tb = $("<table border=1></table>"); for(i=0;i<r.list.length;i++){ tb.append("<tr><td>"+r.list[i].c_id+"</td><td>"+r.list[i].c_name+"</td></tr>"); } $("body").append(tb); } }); }); }) </script> </head> <body> <button id="getjson">获取</button> </body> </html>