[RESOLVED] Help with calculation

fotofx

I am writing a script that calculates number of days to charge for a service. If the customer is only 30 minutes after the day, they only want to charge for 1 day. So 24 hours and 30 minutes will only be 1 day. But 24 hours and 31 minutes will be 2 days. I have an array that has "days", "hours", "minutes" in it already calculated from a calendar. Any thoughts on the easiest way to code this so the "days" will be correct. Essentially I want to round the days up or down depending on the hours but the round point is 30 minutes.

Any help is appreciated.
Steve

bpat1434

So what you're saying is that a customer purchases a service and you want to track the time spent on that service and if it reaches 24 hours, 30 minutes, it rolls to 48 hours. And if it hits 48 hours and 30 minutes, it rolls to 72 hours, etc. etc.

Sounds to me like your best bet might be to take 24 hours * 60 minutes + 30 minutes and modulus that with their time spent. So then if there is a remainder you can just look at it to see how much it is. If it's greater than 30, then it rolls over a day; otherwise, it stays where it is.

For example:

<?php

// Change this to be Hours * 60 + excess minutes
$spent = 48 * 60 + 31; # 2 days, 31 minutes

$max = 24 * 60; // 1440 minutes

$remains = $spent &#37; $max;
$days = floor($spent / $max);

if($remains > 30)
{
    $days++;
}

echo "Spent {$days} day(s) on project.\n";

Hope that helps.

fotofx

Actually it is a reservation system. They dont want to charge for another day unless they are over by more then 30 minutes. I will take a look at what you posted and see if it will work. Thanks

Hmmm... Still a bit confused.. I can convert the days, hours, and minutes to minutes using the following right?

$conv_minutes = (($diff[days] * 1440) + ($diff[hours] * 60) + $diff[minutes]);

But if they are already broken down wouldn't some sort of an if statement work???? I just need to increment (or decrement) the $diff[days] right?

Thanks

fotofx

I have it working with the following.. Using your example.

$conv_minutes = (($diff[days] * 1440) + ($diff[hours] * 60) + $diff[minutes]);
$max = 24 * 60;
$remains = $conv_minutes % $max;
$days = floor($conv_minutes / $max);
if($remains > 30)
{
    $days++;
}

echo $days;

But the only problem now is if the start time and end time are 30 minutes or less. The days comes up as zero. I need it to be 1.

OK, I got it.. Added the following.

if ($days == 0)
{   $days++;
}

bpat1434

Glad you got it working. Don't forget to mark this as resolved if it is.

BTW, I did it by minutes because that's the least common denominator in your question. You to know if each day (1440 minutes) they have passed the 30 minute threshold. Easiest way to compare is with minutes, since 49 hours and 24 minutes is "within" that threshold of 30 minutes past an hour, but after the 48 so it should be 72 hours.

Here's some example code:

<?php
/* Given:
$diff = Array (
     [days] => 2
     [hours] => 2
     [minutes] => 28
);
*/

if($diff['minutes'] > 30)
    $diff['days']++;

// PROBLEM: if hours are > 0, we need to round down, otherwise round up:

if($diff['hours'] != 0 || $diff['minutes'] > 30)
{
    $diff['days']++;
}

May be fewer lines of code; however, in my first example, you know precisely how many minutes have passed and since you know that the first 30 minutes will be 30 and the last 30 minutes will be 1410, you can do a simpler "if > 30" rather than a more complicated "if hours not equal to 0 OR minutes > 30".

Either way will work, I prefer to be explicit in my code with what I'm doing since I may need to rework it later (like if they want to add a grace-period of 30 minutes from noon, I don't need to do much other than do 720 minutes + 30 minutes is the max past the start of a day they can go.