php mysql multiple select, dropdown menu, not saving back to db - php

Question

My dropdown menu is loading from the database, I can select multiple options from the list. When I click my 'Add' button everything appears to be behaving as expected, apart from the fact that nothing is saving back to the database. Unfortunately I've done so many tweaks now, having looked at stack overflow and others, I'm in the realm of confusion rather than progress.
I have set up a logger file to show when a page is reached. So I can confirm that I do get to the insert_rooms.inc.php page.
<?php
session_start(); // gets current session data for user // this function sends headers so it cannot be called after any output.
require_once '../dbconfig.php';
$date = new DateTime();
$date = $date->format("y:m:d h:i:s");
$id = $_SESSION['u_nickname'];
$u_id = $_SESSION['u_id'];
$file = "../ips.txt";
$text = file_get_contents($file);
$text .= $date ." " . $id . " visited insert_rooms.inc.php"."\n"; ///////// this text must be changed for each page
file_put_contents($file, $text);
$conn = OpenCon();
if ($conn)
{
if(isset($_POST['room[]']))
{
$room = mysqli_real_escape_string($conn, $_POST['room']);
if($room) // if a room exists insert into db with users id.
{
foreach($_POST['room'] as $r)
{
//here we need to take for example the word kitchen and find the id belonging to kitchen from the master_rooms table use it to find the master_room_id for kitchen and insert that into the table called user_room_list - along with the user id taken from the session u_id
$room_id = mysqli_query($conn, "SELECT master_room_id FROM master_rooms WHERE master_room_name = '$r'");
$sql = "INSERT INTO user_room_list(user_info_id, master_room_id)
VALUES(' ".$u_id." ', ' " .$room_id. " ') ";
mysqli_query($conn, $sql);
}
}
}
//session_unset();
//session_destroy();
header("Location: ../room_choice.php");
exit();
CloseCon($conn);
}
?>
extract from other file:
<div class="dropdown">
<form action="includes/insert_rooms.inc.php" method="POST">
"Hold down the Ctrl (windows) / Command (Mac) button to select multiple options."
<br/>
<br/>
<select name="room[]" multiple>
<?php
$conn = OpenCon();
if ($conn)
{
$query = "SELECT master_room_name FROM master_rooms";
mysqli_query($conn, $query) or die('Error querying database.');
$result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_array($result, MYSQLI_NUM))
{
for ($i = 0; $i < count($row); $i++)
{
echo "<option value=". $row[$i] . ">" . $row[$i] ."</option>";
}
}
CloseCon($conn);
echo "</select>";
echo " <input type="."submit"." value="." Add "." name="."send"."> ";
echo " </form> ";
}
?>

Answer 1

Below is the solution I found. Code is corrected accordingly.
foreach($room as $r) {
$result = mysqli_query($conn, "SELECT master_room_id FROM master_rooms WHERE master_room_name = '$r' ;");
$room_id_array = mysqli_fetch_array($result);
$room_id = $room_id_array[0]; // required as room_id goes into an array of length 1
$stmt = $conn->prepare("INSERT INTO user_room_list(user_info_id, master_room_id) VALUES (?, ?)");
$stmt->bind_param("si", $u_id, $room_id);
$stmt->execute();