Find out the nearest bigger multiple of (n) PHP - php

Question

I have a requirement where users are forced to choose the multiple of (n) quantity of a product.
The (n) value is set with each product that can be any number.
customer can only purchase the quantity of product in the multiple of (n) quantity set with product.
Suppose if (n) is 5 and user entered quantity as 4 and says Add to Cart. I have to add quantity of that product as 5 automatically.
and if user entered 6 as quantity then I have to add the 10 quantity of that product.
How I go about that?
I am not getting what logic should be applied here.

Answer 1

$entered_quantity = 6;
$suppose_n = 5;
$quantity = ceil($entered_quantity / $suppose_n) * $suppose_n;
echo $quantity;
prints 10

Answer 2

that's not php specific;
what you wonna do is to compute.
ceiling(q / n) * n
where q is the user's quantity,
n is the multiplicity

Answer 3

You could try getting the remainder of the number when dividing by the given n
e.g.:
$n = 5;
$amount = 6; // This would be the input, so replace the 6 with a $_POST/$_GET/etc.
$batches = floor($amount / $n);
$rest = $amount % $n;
if ($rest > 0) {
$batches += 1;
// You could give the user feedback here that they didn't put in a full multiple of $n
}
// $batches now contains the right amount of batches, so to get the total:
$total = $batches * $n;
Ofcourse this can be condensed a lot, but this might give a better overview of what happens :).

Answer 4

Try the below function.
function getNextMultipleOfFive($n) {
$tmp=explode('.',($n/5));
if($tmp[1]) {
return ($tmp[0]+1)*5;
}
return $tmp[0]*5;
}

Answer 5

With a do...while loop:
$q = 6; // quantity by user input
$n = 5; // per purchace amount
$i = 0;
if ($q > 0)
{
do
{
$i += $n;
}
while ($i <= $q);
}
echo $i; // 10
Note: not very effective if $q >> $n