Count number if integers in equation with php - php

Question

I have the need to count the number of integers which i am adding/substracting from equation.
Lets say I have a switch statement like that:
switch($_POST['dni']){
case 2:
$sum1 = $day1;
$sum2 = $day2;
break;
case 3:
$sum1 = $day1 + $day2;
$sum2 = $day3;
break;
case 4:
$sum1 = $day1 + $day2;
$sum2 = $day3 + $day4;
break;
case 5:
$sum1 = $day1 + $day2 + $day3;
$sum2 = $day4 + $day5;
break;
case 6:
$sum1 = $day1 + $day2 + $day3;
$sum2 = $day4 + $day5 + $day6;
break;
case 7:
$sum1 = $day1 + $day2 + $day3 + $day4;
$sum2 = $day5 + $day6 + $day7;
break;
case 8:
$sum1 = $day1 + $day2 + $day3 + $day4;
$sum2 = $day5 + $day6 + $day7 + $day8;
break;
And I have selected $_POST['dni'] to be 7, can I count the number of integers in $sum1 equation?
The things I have tried so far:
explode($sum1, '+') - returns the sum of the equation in array...
count($sum1) - returns 1, because it is counting again the sum of
equation...

Answer 1

Are you thinking of something like this? But without a better example this is all I got for you.
$day1 = 1;
$day2 = 2;
$day3 = 3;
$day4 = 4;
$array = array(
'0' => $day1,
'1' => $day2,
'2' => $day3,
'3' => $day4
);
$sum = array_sum($array);
echo 'Sum: = ' . $sum . '<br>'; //Outputs 10
echo '# of Integers: = ' . count($array); //Outputs 4

Answer 2

So I think it's a bad practice to use switch statement and re-did your code by -infer- method and resolved the issue, I hope it helps:
<?php
// In my example I have to declare the days and sums variables
// but I'm assuming you have them declared in your code already…
$dni = 8; // Or $_POST['dni'];
$day1 = 3;
$day2 = 4;
$day3 = 6;
$day4 = 8;
$day5 = 2;
$day6 = 5;
$day7 = 3;
$day8 = 10;
$day9 = 15;
$sum1 = $sum2 = 0;
// Here by what I saw in your code, can be simplified with basic arithmetic.
// First I divide $dni by 2, if it's an odd number it'll result in a float number.
// By using ceil we round it to the closest even number.
$primary_sum = ceil($dni / 2);
// Then just subtract the $primary_sum from $dni
// and you'll have your $secondary_sum.
$secondary_sum = $dni - $primary_sum;
// Here what I did is make an array with $dni number of keys.
// (Serves as your switch statement).
foreach (array_fill(1, $dni, '') as $day_number => $x)
{
// Follow the logic.
if ($day_number <= $primary_sum)
{
// This is a little trick but works like a charm.
$sum1 += ${"day{$day_number}"}; // It's basically $day1... and on.
}
else if ($day_number > $primary_sum)
{
// This is a little trick but works like a charm.
$sum2 += ${"day{$day_number}"};
}
}
// Number of integers in first sum (sum1)
var_dump($primary_sum);
// Number of integers in second sum (sum2)
var_dump($secondary_sum);
// The total sum of sum1 (pretty obvious).
var_dump($sum1);
// The total sum of sum2 (pretty obvious).
var_dump($sum2);
You can paste that code in phpsandbox and see it in action.