i want to access mysql database table on given conditions in drop down menu [on hold]
- by user3909877
as the code below is accesing the database table directly but i want it to display the table content on giving conditions in drop down menu like when i select islamabad in one drop down menu and lahore in other as given in code and press search buttonn then it display the table flights.but it is displaying it directly
<p class="h2">Quick Search</p>
<div class="sb2_opts">
<p>
</p>
<form method="post" action="">
<p>Enter your source and destination.</p>
<p>
From:</p>
<select name="from">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<p>
To:</p>
<select name="To">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<input type="submit" value="search" />
</form>
</form> </table>
<?php
$from = isset($_POST['from'])?$_POST['from']:'';
$to = isset($_POST['to'])?$_POST['to']:'';
if( $from =='Islamabad'){
if($to == 'Lahore'){
$db_host = 'localhost';
$db_user = 'root';
$database = 'homedb';
$table = 'flights';
if (!mysql_connect($db_host, $db_user))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
die("Query to show fields from table failed");
}
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h1>Table: {$table}</h1>";
echo "<table border='1'><tr>";
while($row = mysql_fetch_row($result))
{
echo "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>\n";
}
}
}
mysqli_close($con);
?>
