Php SQL insert issue

id_file
id_cust - foreign key for idcust from table Customers
register_date
id_object - foreign key for idcust from table Objects

idcust
cust_name
address
phone

idobject
name_object

<pre><form method="post" action="">
<tr>
<td>Customer name: </td> <td> 
<select name="cust_name">
<?php 
$sql = mysqli_query($conn, "SELECT * FROM customers");
while ($row = $sql->fetch_assoc()){ 
echo "<option value=\"cust_name1\">" . $row['cust_name'] . "</option>";
}?>
</select></td></tr>

<tr>
<td>Register date</td>
<td> <input  type="date" name="register_date"/></td>
</tr>

<tr>
<td>Object: </td> <td> 
<select name="object">
<?php 
$sql = mysqli_query($conn, "SELECT * FROM objects");
while ($row = $sql->fetch_assoc()){ 
echo "<option value=\"name_object1\">" . $row['name_object'] . "</option>";
}?>
</select></td></tr>
<tr><td colspan=2> <input name ="submit" type="submit" value="Add a new file"></td>
</tr></form>

<?php

 $conn = mysqli_connect("localhost", "root", "", "testdb");
 if ($conn -> connect_error){
   die("Connection failed:". $conn-> connect_error);
 }
 if(!empty($_POST['submit'])){

   $cust_name = $_POST['cust_name'];
       $register_date  = date('Y-m-d',strtotime($_POST['register_date']));
       $name_object =$_POST['name_object'];

   $sql = "INSERT INTO files VALUES (
   (SELECT * FROM customers c WHERE c.cust_name = $cust_name),
   $register_date,
   (SELECT * FROM objects o  WHERE o.name_object = $name_object)
   )";

   $conn->query($sql);
   if($conn->error){
     echo $conn->error;
   } else
   {
     $message= "We have added the file no. " .$conn->insert_id;
   }
 }
 ?>