我jquery ajax 取json数据总是不能显示

public void ProcessRequest(HttpContext context)
  {
  //context.Response.ContentType = "text/plain";
  //不让浏览器缓存
  context.Response.ContentType = "application/json";
  context.Response.Charset = "utf-8";

  string result = "";

  //string data = "[{name:\"fan\",age:26},{name:\"wang\",age:25}]";
  SqlConnection con = new SqlConnection(System.Configuration.ConfigurationManager.ConnectionStrings["CookbookConnectionString"].ConnectionString);


  SqlDataAdapter sda = new SqlDataAdapter("SELECT * FROM cc", con);

  DataTable dt = new DataTable();
  // dt.TableName = "cc";
  string cc = "cc";
  sda.Fill(dt);
  result = DataTableToJSON(dt,cc);

  context.Response.Write(result);
  }

DataTableToJSON（dt,cc)json数据为{"cc":[{"id":"1","name":"新闻"},{"id":"2","name":"今日要闻"},{"id":"3","name":"焦点新闻"},{"id":"4","name":"今日精彩"},{"id":"5","name":"时事要闻"},{"id":"6","name":"蛋疼新闻"},{"id":"7","name":"Oracle"},{"id":"8","name":"mssql"},{"id":"9","name":"Dba"},{"id":"10","name":"Dick"}]} 

前台代码
jquery代码
<script>
 $.getJSON("Handler.ashx", function(json) 
  { var json = eval("(" + json.d + ")");
  var xx = json.cc;
   
  //alert(""+json1+"");
  $.each(xx,function(i,entry)
  {  
  alert(data[i].name);  
 

$('#a').append("<option value='"+entry.name+"'>"+entry.name+"</option>"); })
});
</script>


数据怎么都绑定不到select控件a中

求高人指点