将一个表中的更新信息插入另一个表

Question

我有两个桌子。一个叫做雇员，另一个叫做请求。我将表雇员中雇员的更新数据放在表请求中。我想从URL中获取一个雇员的ID，然后从请求表中获取有关该雇员的信息，并将其插入到表employee中。但是，它不起作用，也没有放置错误。我假设问题出在从URL获取ID。这是代码

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "project";

$conn =  new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

mysqli_select_db($conn, 'project');


//$id = $_GET['Emp_ID'];
$id = isset($_GET['Emp_ID']) ? $_GET['Emp_ID'] : '';
$sql = "SELECT * from requests where Emp_ID = '$id'";
$records = mysqli_query($conn, $sql) or die (mysqli_error($conn));




while($row = mysqli_fetch_array($records)){

  $LastName = $row['LastName'];
  $FirstName = $row['FirstName'];
  $Age = $row['Age'];
  $salary = $row['salary'];
  $email = $row['email'];
  $password = $row['emp_password'];
  $address = $row['address'];
  $telephone = $row['telephone'];

  $sql = "UPDATE `employees` SET `LastName`='$LastName',`FirstName`='$FirstName',`Age`='$Age',
  `salary`='$salary',`email`='$email',`emp_password`='$password',`address`='$address',`telephone`= '$telephone'  WHERE Emp_ID = '$id'";



}
 $sql = "DELETE from requests where Emp_ID = '$id'";

?>