ajax后台处理返回json值示例代码


            复制代码 代码如下: 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; }