ajax返回数据不跳success

<form action="" onsubmit="post_data();return false;" id="formT" method="post" class="form-T">

 

form标签内加 onsubmit="post_data();return false;"

 

function post_data(){

$.ajax({

                url: "",

                type: "POST",

                data: form,

                dataType: "json",             

                success: function (res) {

                alert(res.msg);

                },

                error:function(){

                    console.log("请求失败")

                },

                timeout:2000,

                });

                return false;

            })

 }

函数采用function post_data()写法

$.ajax添加return false;