ajax后台处理返回json值

  public ActionForward xsearch(ActionMapping mapping, ActionForm form,HttpServletRequest request, HttpServletResponse response)throws Exception {String parentId = request.getParameter("parentId");String supplier = request.getParameter("supplier");List itemList = new ArrayList();if(parentId.equals("")){parentId="0";}Map map=new TawApTreeServlet().getTypeList(parentId, supplier);for (Iterator rowIt = map.keySet().iterator(); rowIt.hasNext();) {String id = (String) rowIt.next();TawCommonsUIListItem uiitem = new TawCommonsUIListItem();uiitem.setItemId(id);uiitem.setText((String)map.get(id));uiitem.setValue(id);itemList.add(uiitem);}response.setContentType("text/xml;charset=UTF-8");// 返回JSON对象response.getWriter().print(JSONUtil.list2JSON(itemList));return null;}