更新代码在PHP中不适用于jquery Ajax？

我有4张独立的桌子——螺柱、国家硕士、学术硕士、国家硕士、城市硕士。我正在尝试更新，但它会打开模式并显示所有获取的值，然后当我更改值并单击“更新”时，模式会关闭，数据不会更新。你能说我做错了什么吗？先谢谢你。 我的更新按钮看起来像这样

<button type="submit" name="update" class="btn btn-info">Update</button>

和update.php页面

<?php
include("connection.php");
    $no=$_POST['stud_no'];
    $name=trim($_POST['name']);
    $mob=trim($_POST['mob_no']);
    $dob=trim($_POST['dob']);
    $add=trim($_POST['add']);
    $photo=trim($_FILES['photo']['name']);
    $gen=trim($_POST['gender']);
    $cn=trim($_POST['country']);
    $st=trim($_POST['state']);
    $ct=trim($_POST['city']);

if(isset($_POST['stud_no'])){
$id= $_POST["stud_no"];
$qry="select * from stud s, country_master_academic c,master_state st, master_city ct where s.country=c.country_code and s.state=st.state_code and s.city=ct.city_code and stud_no='".$id."'";
$res = mysqli_query($conn,$qry);
$row=mysqli_fetch_array($res);
echo json_encode($row);
}
else
{
    $qry="update stud set stud_name='$name',mobile='$mob',dob='$dob',address='$add',gender='$gen',country='$cn',state='$st',city='$ct' where stud_no='$no'";
    $data=mysqli_query($conn,$qry);
    if($data)
    {
        echo '<script language="javascript">';
        echo 'alert("Updated Successfully")';
        echo '</script>';
    }
    else {
        echo '<script language="javascript">';
        echo 'alert("Cannot update record")';
        echo '</script>';
    }
    if(!empty($_FILES['photo']['name'])){
    $qry1= "update stud set photo='$photo' where stud_no='$no'";
    $data1=mysqli_query($conn,$qry1);

    if($data1){
        $target_dir="images/";
            $target_file=$target_dir.basename($_FILES["photo"]["name"]);
            $imageFileType=pathinfo($target_file,PATHINFO_EXTENSION);
            if(move_uploaded_file($_FILES["photo"]["tmp_name"],$target_file)){
                echo '<script language="javascript">';
                echo 'alert("Image upload successfully")';
                echo '</script>';

            }  else {
                echo '<script language="javascript">';
                echo 'alert("Cannot Upload")';
                echo '</script>';
            }
    }
}
}

html：

<form id="FormSubmit" method="post">

@Yogendrasinh我在这里问了一个新问题that@BhoomiPatel看到这位女士了吗
<form id="FormSubmit" method="post">

$(document).ready(function(){
     $( "#FormSubmit" ).submit(function( event ) {
        event.preventDefault();
        $.ajax ({
            url: "update.php",
            type: "POST",
            data:  new FormData(this),
            contentType: false,
            cache: false,
            processData:false,
            success: function(response) {
        console.log(response);  
                        }
                    });
                });
                });