How should I write this specific condition in php? - php

Question

The Challenge is:
If the book is returned on or before the expected return date, no
fine will be charged (i.e.: fine=0).
If the book is returned after the expected return day but still
within the same calendar month and year as the expected return date,
fine=15 Hackos × (the number of days late).
If the book is returned after the expected return month but still
within the same calendar year as the expected return date, the
fine=500 Hackos × (the number of months late).
If the book is returned after the calendar year in which it was
expected, there is a fixed fine of 10000 Hackos.
And My code is:
<?php
$expectedDay = "6";
$expectedMonth = "6";
$expectedYear = "2015";
$returnDay = "9";
$returnMonth = "6";
$returnYear = "2015";
if ($expectedDay >= $returnDay && $expectedMonth >= $returnMonth && $expectedYear >= $returnYear) {
echo "Fine = 0";
}elseif ($expectedDay < $returnDay && $expectedMonth == $returnMonth && $expectedYear == $returnYear) {
$fine = 15 * ($returnDay-$expectedDay);
echo "Fine = ".$fine;
}elseif (($expectedDay <= $returnDay || $expectedDay >= $returnDay) && $expectedMonth < $returnMonth && $expectedYear == $returnYear) {
$fine = 500 * ($returnMonth-$expectedMonth);
echo "Fine = ".$fine;
}else{
echo "Fine = 1000";
}
?>
Its running well.But failed when the input is:
$expectedDay = "28";
$expectedMonth = "2";
$expectedYear = "2015";
$returnDay = "15";
$returnMonth = "4";
$returnYear = "2015";
How do I write for this condition?
Note: This is not a business logic. It is just Practice purposes. I am a beginner in PHP.

Answer 1

Something along these lines:
function calculateLateFees(DateTime $deadline, DateTime $returned) {
if ($returned <= $deadline) {
return 0;
}
if ($returned->format('Y') > $deadline->format('Y')) {
return 10000;
}
if ($returned->format('n') > $deadline->format('n')) {
return ($returned->format('n') - $deadline->format('n')) * 500;
}
return $deadline->diff($returned)->days * 15;
}
$deadline = new DateTime('2015-02-28');
$returned = new DateTime('2015-04-15');
echo calculateLateFees($deadline, $returned), ' Hackos';

Answer 2

You do not have to compare dates when the month condition is checked.
<?php
$expectedDay = "6";
$expectedMonth = "6";
$expectedYear = "2015";
$returnDay = "9";
$returnMonth = "6";
$returnYear = "2015";
$returnDate = new DateTime($returnDay.'-'.$returnMonth.'-'.$returnYear);
$expectedDate = new DateTime($expectedDay.'-'.$expectedMonth.'-'.$expectedYear);
if ($returnDate <= $expectedDate) {
echo "Fine = 0";
}elseif ($expectedDay < $returnDay && $expectedMonth == $returnMonth && $expectedYear == $returnYear) {
$fine = 15 * ($returnDay-$expectedDay);
echo "Fine = ".$fine;
}elseif ($expectedMonth < $returnMonth && $expectedYear == $returnYear) {
$fine = 500 * ($returnMonth-$expectedMonth);
echo "Fine = ".$fine;
}else{
echo "Fine = 1000";
}
?>
Try that.

Answer 3

First, I must say that this is a very unfair calculation of fines. What if the expected day is 31.12.2016 and the book was returned 01.01.2017? According to your method the a fine of 10000 will be imposed for the lapse of one day just because it was across the margin of two separate years.
I would recommend you to calculate the fine according to the number of days late.
$lateDays = date_diff($expectedDate, $returnDate, false);
if ($lateDays > 0) {
if ($lateDays < 30)
$fine = 15 * $lateDays
else
if ($lateDays > 365)
$fine = 10000
else
$fine = 500 * $lateDays / 30
}

Answer 4

if($returnYear > $expectedYear){
$fine = 1000;
}
else{
if($returnMonth > $expectedMonth ){
$fine = 500 * ($returnMonth-$expectedMonth);
}
else{
if($returnDay>$expectedDay){
$fine = $returnDay - $expectedDay;
}
else{
$fine = 0;
}
}
}
echo $fine;
Above logic is built taken into consideration that months in number crossed will take complete month cost example :
if expected return date is 28/04/2016 and actual return date is 02/05/2016 then since month is changed the below algorithm takes fine #complete month.
If you want exact days difference to be considered for months/year calculation then we can write a different logic all together

Answer 5

<?php
$expectedDay = "6";
$expectedMonth = "6";
$expectedYear = "2015";
$returnDay = "9";
$returnMonth = "6";
$returnYear = "2015";
$expectedate=$expectedYear.'-'.$expectedMonth.'-'.$expectedDay;
$returndate=$returnYear.'-'.$returnMonth.'-'.$returnDay;
$expected=date_create($expectedate);
$return=date_create($returndate);
$interval=date_diff($expected, $return);
$valor=$interval->format('%R%a');
if ($valor>0) {
if ($returnMonth==$expectedMonth && $returnYear==$expectedYear) echo "Fine=".(15*$valor);
if ($returnMonth!=$expectedMonth && $returnYear==$expectedYear) echo "Fine=".(500*($returnMonth-$expectedMonth));
if ($returnYear!=$expectedYear) echo "Fine=1000";
} else echo "Fine=0";
?>