关于如何接受异步ajax请求返回前台的数据
1、第一种方式通过bean,后台需封装成json数据格式,前台paseJSON(); public void checkCode() { ResponseMessage rm = new ResponseMessage(true,null); String hql = "from TblWrPeople t where t.wrPeopleCode='"+code+"' and t.delFlag='"+Constants.DELFLAGA+"'"; List<TblWrPeople> tblWrPeopleList = (List<TblWrPeople>)this.workResService.find(hql); if (tblWrPeopleList != null && tblWrPeopleList.size() > 0) { rm.setSuccess(false); rm.setErrorMsg("数据库中已存在该代号,请重新录入"); this.responseAJAX(JSONObject.fromObject(rm).toString(),getResponse()); } } 前台获取: function checkCode() { var code = $("#wrPeopleCode").val(); $.post('workRes_checkCode.action',{"code":code},function(result){ result = $.parseJSON(result); if (result.success == false){ alert(result.errorMsg); $("#wrPeopleCode").val(''); } }); } 2、后台直接返回一个String类型的字符串,前台的话不再需要parseJSON,直接获取result即可; (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |