updating a field in mysql using php - php

Question

|room type|price|avroom|
|standard |50 |20 |
| super |60 | 7 |
|suite |80 | 4 |
if reservation is made for "room type" = standard
the value entered in the "no of rooms" field in the form should be subtracted from the value in the "avroom" for the "room type" on the database
the code below outputs the form
</select></td>
<th align=left>ROOM TYPE :</th>
<td>
<?php
echo "<select name=txttype>";
$qup="select * from tariff where avroom > 0";
$rs=mysql_query($qup);
while($res=mysql_fetch_row($rs))
{
echo "<option value='".$res[0]."'>".$res[0]."</option>";
}
echo "<select>";
echo "</td>";
?>
<tr>
<th align=left>NO OF ROOMS :</th>
<td><select name=txtroom>
<?php
for($i=1;$i<=20;$i++)
{
echo "<option value=$i>$i</option>";
}
?>
please if you need any more info let me know

Answer 1

First thing I noticed in your mysql table is there cannot have a column like "room type" in a table. It should be edited to "room_type".
First of all, if you only need to display the room type in the top select box, your sql should be edited to $qup="select room_type from tariff where avroom > 0";. Then post the selected values using a html form.
From the other end where the form is posted, try something like,
$room_type= $_POST['txttype'];
$rooms = mysql_real_escape_string($_POST['txtroom']);
if($room_type == "standard")
{
$query = "UPDATE tariff SET avroom=avroom-{$rooms} WHERE room_type='standard'";
// Now execute the above query here.
}