PHP Counting Negative Numbers and Bold Only Even Numbers - php

Question

On a form, I used PHP to display every number from 1 to the entered number. For example, if I enter 10 on the form, it displays 1 2 3 4 5 6 7 8 9 10. Now, I want it to be able to handle negative numbers by counting up to 0 (-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0) and make every even number in the results bold, e.g., 2 4 6 8 10, etc.). I've searched exhaustively for the answers without any luck. How would any of you suggest doing this? My code for the first part is displayed below. Thank you in advance.
<?php
$num = $_POST['num'];
$limit = $_POST['num'];
echo "<pre>";
do {
echo ($counter).'<br>';
$counter++;
} while ($counter <= $limit);
echo "<pre>";
?>

Answer 1

You can do this
<?php
$num = $_POST['num'];
$limit = $_POST['num'];
echo "<pre>";
do {
if( $counter % 2 == 0 )
{
echo "<strong>" . $counter . "</strong><br />";
}
else
{
echo ($counter).'<br>';
}
$counter++;
} while ($counter <= $limit);
echo "<pre>";
?>
Solution 1
Ok, some more information about the %. This is a modulo. It gives you back the remaining number if you divide it by the mod number. For example
0 % 2 = 0
1 % 2 = 1
2 % 2 = 0
3 % 2 = 1 this because 3 / 2 = 1 and a bit,
you can't divide the last 1 completely by 2. So remain 1
4 % 2 = 0
Solution 2
Like crush said, you can use $counter & 1. What does this do?
If you look at a number bitwise. You want to AND it wit 1.
Bitwise number 2 = 0010 AND it with 0001 and your return will be 0000 (zero).
Bitwise number 3 = 0011 AND it with 0001 and your result will be 0001 (one).
If you ceep that going and only check the last bit, you can see if it is a even number. More about bitwise operations.