通过代码示例分析给大家介绍ajax实现页面局部跳转与结果返回，具体内容如下：

<input type="button" class="btn" value="提报" name="submit4" onClick="tibao();">
function tibao(){var id='';var URL = <select:link page="/smokeplan.do?method=Tibao&idset="/>+id;  $.ajax({url: URL,     　　type: 'GET',    　　success: function(result) {           alert(result);        }  });}
后台Java类处理过程为：
//提报    public void Tibao(ActionMapping mapping, ActionForm form,        HttpServletRequest request, HttpServletResponse response) throws Exception {      String idset=request.getParameter("idset");      CallHelper helper = initializeCallHelper("L_SmokeBoxtibaoWLDan", form,request, false);      helper.setParam("bill_ids",idset);      helper.setParam("personid",getPersonId(request));      helper.execute();      PrintWriter write = response.getWriter();      write.print(helper.getOutput("message"));      write.close();    }
这里做的操作的结果是反映到response对应的位置，于是拿到属于response的流，而不是new一个出来。
使用AJAX如何实现页面跳转
function doGuahao(){ if(checkdata()) { if(document.form1.result_flag.value=="0") {  return false; } else {  if(document.form1.checktype.value=="danganhao")  {  form1.action = "<%=formAction%>";  form1.submit();  }  if(document.form1.checktype.value=="xingming")  {  form1.action = parent.left.url2;  form1.submit();  }  if(document.form1.checktype.value=="shenfenzheng")  {  form1.action = "<%=formAction%>";  form1.submit();  } } } }