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;

	}