Problem trying to add multiple choose to mysql - php

Question

I’m facing a big problem here. My hotel company purchased a Booking system made with PHP from a South African company, their contract expired now they are now charging a lot of money to update the contract and add new functions that we need, so now all the burden is left for me… etc. So I have this table:
and I want to add a multiple type of room at each room. E.g.: Room 1, Type: Simple, Two Beds, Three Beds
Table coode:
<div class="table-responsive">
<table id="datatable-fixed-header" class="table datatable-show-all table-striped table-bordered">
<thead>
<tr>
<th>Nº</th>
<th>Floar</th>
<th>Room No.</th>
<th>Price per Night ( In <?php echo $currency_symbol; ?> )</th>
<th>Price per Week ( In <?php echo $currency_symbol; ?> )</th>
<th>Type</th>
<th>State</th>
<th>Action</th>
</tr>
</thead>
<tbody>
<?php
if (count($room_details) > 0) {
$s_no = 1;
foreach ($room_details as $room) {
echo'<tr>
<td>' . $s_no . '</td>
<td>' . $room["property_name"] . '</th>
<td>' . $room["room_number"] . '</th>
<td class="text-right format_currency_td">' . $room["room_standard_night_rate"] . '</td>
<td class="text-right format_currency_td">' . $room["room_standard_weekly_rate"] . '</td>
<td>' . $room["room_type_name"] . '</td>';
if ($room["room_status"] == "AVAILABLE")
echo'<td><span class="label label-success">' . $room["room_status"] . '</span></td>';
else if ($room["room_status"] == "OCUPADO")
echo'<td><span class="label label-info">' . $room["room_status"] . '</span></td>';
else if ($room["room_status"] == "MANUNTEINCE")
echo'<td><span class="label label-warning">' . $room["room_status"] . '</span></th>';
else if ($room["room_status"] == "OUT OF SERVICE")
echo'<td><span class="label label-danger">' . $room["room_status"] . '</span></td>';
echo'<td>';
if ($room["room_status"] != "OUT OF SERVICE")
echo'<div class="th_options_kit">
<span class="th_options btn bg-teal-400 btn-icon btn-rounded btn_edit_room" data-room="' . $room["room_id"] . '" data-popup="tooltip" title="Edit Room" data-placement="left"><i class="icon-pen"></i></span>
<span class="btn bg-danger-400 btn-icon btn-rounded btn_delete_room" data-room="' . $room["room_id"] . '" data-popup="tooltip" title="Delete Room" data-placement="left"><i class="icon-bin"></i></span>
</div>';
else if ($room["room_status"] == "OUT OF SERVICE")
echo'<div class="th_options_kit">
<span class="th_options btn bg-teal-400 btn-icon btn-rounded btn_edit_room" data-room="' . $room["room_id"] . '" data-popup="tooltip" title="Room is out of service" data-placement="left"><i class="icon-pen"></i></span>
</div>';
echo'</td>
</tr>';
$s_no++;
}
}else {
}
?>
</tbody>
</table>
</div>
Php Code:
public function getRoom($where = FALSE) {
$this->db->select('room.`room_id`, '
. 'room.`property_id`, '
. 'room.`room_number`, '
. 'room.`room_type`,'
. 'room.`room_standard_night_rate`,'
. 'room.`room_standard_weekly_rate`,'
. 'room.`room_status`,'
. 'prop.`property_name`,'
. 'room_type.`room_type_name`'
);
$this->db->from('hm_room room');
$this->db->join('hm_room_type room_type', 'room.`room_type` IN (room_type.`room_type_id`)');
$this->db->join('hm_property prop', 'room.`property_id` IN (prop.`property_id`)');
if ($where !== FALSE) {
foreach ($where as $whr) {
$this->db->where_in($whr["column_name"], $whr["data"]);
}
}
$this->db->order_by("prop.`property_name`", "ASC");
$this->db->order_by("room.`room_number`", "ASC");
$query = $this->db->get();
return $query->result_array();
}
Help me solve this please.

Answer 1

I hope i understand your question, your table is currently just showing Room Type e.g. Simple. Is what you want it to show now "Simple, Two Bed" instead of just "Simple" ?
Then what you need to do is concatenate your fields. I can't seem to see what the database field for "Two Bed", but assuming it's "room_beds", then you need to do the below:
First step is to add the database field name to the query, assuming the name of the field in the database is called "room_beds"
$this->db->select('room.`room_id`, '
. 'room.`property_id`, '
. 'room.`room_number`, '
. 'room.`room_type`,'
. 'room.`room_standard_night_rate`,'
. 'room.`room_standard_weekly_rate`,'
. 'room.`room_status`,'
. 'prop.`property_name`,'
. 'prop.`room_beds`,' -- Add the here
. 'room_type.`room_type_name`');
Then finally you need to concatenate the room_beds field to the room_type_name that you already have as below:
<td>'. $room["room_type_name"] .' ,'. $room["room_beds"] .'</td>';