More efficient rounding calculator

[Adamhumbug]

Hi All,

I have a rental system - if you rent for 3 or more days over a week you pay for 2 weeks, 3 or more days over 2 weeks and you pay for 3 and so on.

As i need to be able to see how many weeks someone is renting equipment for i have built a js function but wondered if there was a more effience way of getting the weeks value than the following.

            if ($days == 0){
                $weeks = 0
            }else if ($days < 10) {
                $weeks = 1
            } else if ($days >= 10 && $days < 16) {
                $weeks = 2
            } else if ($days >= 17 && $days < 23) {
                $weeks = 3
            } else if ($days >= 24 && $days < 30) {
                $weeks = 4
            }

Thanks as always

Edited August 18, 2023 by Adamhumbug

[Barand]

Try

$weeks = intdiv($days, 7) + ($days%7 > 3 ? 1 : 0);

 

[Adamhumbug]

8 minutes ago, Barand said:

Try

$weeks = intdiv($days, 7) + ($days%7 > 3 ? 1 : 0);

 

Amazing - as ever!!

[Barand]

Correction:

> 3 should be >= 3

[Adamhumbug]

Sorry my testing was wrong here - i am gettin intdiv is undefined.

 

Looking into it - will report back once tested properly.

[Adamhumbug]

Ah, this is a PHP function?  I am in JS at the mo

[Barand]

The code you posted was using PHP variable names. ($days, $weeks)

intdiv (x, y) == floor(x/y)

[Adamhumbug]

19 minutes ago, Barand said:

The code you posted was using PHP variable names. ($days, $weeks)

intdiv (x, y) == floor(x/y)

Apologies, i am using JQuery.

 $weeks = Math.floor($days/7) + ($days%7 >= 3 ? 1 : 0);

The above is working beautifully.