Question

java域名为项目名加servlet的名字加参数跳转

在提交页面提交数据后，像下面这样，然后跳到另一个地址去，要怎么做到？
http://localhost:8080/yftpay/bankPayServlet?&summary=oppo&amount=0.01&tradeNo=20180427112839&sign=5C66C6B2DEC698309DB995F8EC350D85&tradeDate=20180427&version=1.0.0.0&bankId=CMB&expireTime=&service=TRADE.B2C&extra=test&clientIp=110.87.91.110¬ifyUrl=http://localhost:8080/yftpay/notifyServlet&merId=2017122011010020

Answer 1

$.ajax({
        type: "post",//方法类型
        url:"./A.action",
        data:$('#from').serialize(),// 序列化表单值  
        contentType:'application/x-www-form-urlencoded;charset=utf-8',
        dataType: "json",
        success: function(data){
            if (data.result == false) {
              layer.msg("注册失败!", {time:1000});
             return false;
            }else if(data.result){
              layer.msg('注册成功',{time:500},function() {
                  window.location.href='./B.action';//注册成功直接跳B页面
                })
            }
        },
        error:function() {
            layer.msg("异常！");
        },
        end: function () {
                location.reload();
            }
    },"json");

Answer 2

在web.xml里配置servlet的url