sql name(age)...

<form name = "zad" method = "POST">
<select name = "wybor">
<option value = "NAME">NAME</option>
<option value = "AGE">AGE</option>
<option value = "SALARY">SALARY</option>
</select>
<input type="submit" />
</form>

<?php
class MyDB extends SQLite3
{
function __construct()
{
$this->open('company.dat');
}
}
$wybor = $_POST['wybor']; //kryterium sortowania wybrane w select

$db = new MyDB();//zakladam ze baza jest juz wypelniona danymi

$result = $db->query('SELECT * FROM COMPANY ORDER BY '.$wybor.';');
while($row = $result->fetchArray()){
echo $row['NAME']."(".$row['AGE']."),".$row['ADDRESS'].";".$row['SALARY']." $<br>";
}
$db->close();

?>