mysqli update Query有什么问题吗?

我无法以表格形式更新我的CRUD web应用程序中的现有数据。这个查询有什么问题吗?这是我的参考来源，我已经完全按照这里的UPDATE查询INSERT, UPDATE和DELETE与mysqli。这是我的代码。

<?php
//error_reporting(E_ALL^E_NOTICE);
    function chgDate($date){
    $temp=explode("-",$date);
    return $temp[2]."-".$temp[1]."-".$temp[0];
}
$json=array();

$ic = $_POST['IC'];
$Fic = $_POST['fromIC'];
$name = $_POST['formName'];
$tel = $_POST['formTelephone'];
$gender = $_POST['formGender'];
$email = $_POST['formEmail'];
if(isset($_POST['formUni'])){
    $uni = $_POST['formUni'];
    }
$age = $_POST['formAge'];
$address = $_POST['formAddress'];
$dob = $_POST['formDOB'];
$process= $_POST['process'];
//include ("connect_db.php");
//include_once('connect_db.php');
$db = mysqli_connect("localhost","root","admin","li") or die("Connection Error: " . mysqli_error());
if($process == 'save'){
    $SQL="Insert into biodata (IC, Name, Telephone, Gender, Email, University, Age, Address, DOB) values ('$Fic', '$name', '$tel', '$gender', '$email', '$uni', '$age', '$address', '".chgDate ($dob)."')";
    $json['newrow']=$Fic;
} else if ($process == 'edit') {
    $SQL="UPDATE biodata SET IC='$Fic', Name='$name',  Telephone='$tel', Gender='$gender', Email='$email', University='$uni', Age='$age', Address='$address, DOB ='".chgDate ($dob)."' WHERE IC= '$ic'";
} else if ($process == 'delete') {
    $SQL = "DELETE FROM biodata WHERE IC='$ic'";
}
$data = mysqli_query($db, $SQL);
if($data){
    $json['msg']='success';
}else{
    $json['msg']='fail';
}
echo json_encode($json);

?>