java - jsp页面通过ajax发送到servlet，反之亦然

Question

我通过ajax从jsp页面向servlet发送orderid，如果订单表的orderid不匹配，则会显示错误消息在同一 jsp 页面中，通过该消息可以很好地发出请求。现在我的问题是当我发送有效的 orderid 时它应该通过servlet将页面转发到另一个jsp页面，该页面没有发生并且也没有显示任何错误

jsp页面

  <input type="hidden" name=cmd  value="single"/>
    Enter Order No <input type="text" name="oid" id="oidss" /><input type="button" value="Search" id="finduy" />

jquery代码

$('#finduy').click(function(event){

            $.ajax({
                type:'POST',
                url:"Order",
                style:"full",
                maxRows:12,
                dataType:"json",
                data:{"cmd":"single","oid":encodeURIComponent($('#oidss').val())},

                beforeSend:function(){/*alert("data is sending")*/},
                //problem is here
                success:function(data,textStatus,jqXHR){
                        if(data.success){

                }
                else
                    {
                    alert("invalid Order No");
                    $('#oidss').val("");
                    $('#oidss').focus();


                    }

                },
                error:function(jqXHR, textStatus, errorThrown){
                console.log('textStatus:' + textStatus);
                console.log('errorThrown:' + errorThrown);
                 console.log("Something really bad happened " + textStatus);
                 console.log("jqXHR.responseText " +jqXHR.responseText);

                  $("#ajaxResponse").html(jqXHR.responseText);
            }
            });

servlet 代码

if(!(cmd==null)&&cmd.equalsIgnoreCase("single")){

        OrderDB odb = new OrderDB();
        Order orders =  new Order() ;
        List<OrderDetail> odetaillist = new ArrayList<OrderDetail>();
        List<Order> ordercustlist = new ArrayList<Order>();


        String orderids = request.getParameter("oid");


        try {
            orders = odb.getOrdersById(orderids);
            odetaillist = odb.getOrdersDetailsByOrderId(orderids);
            ordercustlist = odb.getOrderFromCustomer(orderids);
            Customer customerinfo = odb.getOrderCustomer(orderids);
            request.setAttribute("OrderSingle", orders);
            request.setAttribute("OrderDetail", odetaillist);
            request.setAttribute("OrderCust", ordercustlist);
            request.setAttribute("CustomerInfo", customerinfo);
            redirect=orderinfo;
            RequestDispatcher view = request.getRequestDispatcher(redirect);
            view.forward(request, response);

            if(orders.getOrderid() == null){
                JsonObject myobj =  new JsonObject();
                myobj.addProperty("success", false);
                PrintWriter out = response.getWriter();
                out.print(myobj);
                out.close();
                return;
            }



        } catch (SQLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

Answer 1

您始终可以从 servlet 发回响应代码，然后执行转发。例如，您可以发送成功，在这种情况下您可以将 document.location 发送到新的 servlet 位置。

现在还可以使用 ajax 在 html5 中上传文件。