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;
	}