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;

}