ajax后台处理返回json值示例代码

beyond16548 22 0 PDF 2020-12-03 05:12:57

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

用户评论
请输入评论内容
评分:
暂无评论