将数组中的元素按两个对分配，然后找出差值和总和

Let's say I have this array

$number = [2,1,4,3,6,2];

1. First pair the elements on an array by two's and find their difference so this is the output in the first requirement

   $diff[] = [1,1,4];
   

2. Second sum all the difference this is the final output

   $sum[] = [6];
   

Conditions:

1. the array size is always even
2. the first element in a pair is always greater than the second one, so their is no negative difference

What I've done so far is just counting the size of an array then after that I don't know how to pair them by two's. T_T

Is this possible in php? Is there a built in function to do it?

One line:

$number = [2,1,4,3,6,2];
$total = array_sum(array_map(function ($array) {
    return current($array) - next($array); 
}, array_chunk($number, 2)));
echo $total;

Well with your conditions in mind I came to the following

$number = [2,1,4,3,6,2];
$total = 0;
for($i = 0; $i < count($number); $i+=2) {
    $total += $number[$i] - $number[$i + 1];
}

This should work fine:

<?
$number = array(2,1,4,3,6,2);
for($i=0;$i<count($number); $i+=2){
    $dif[] = $number[$i] - $number[$i+1];
}
print_r($dif);
$sum = 0;
foreach ($dif as $item){
    $sum += $item;
}
echo 'SUM = '.$sum;
?>

Working CODE

If you want all the different stages kept,

$numbers = [2,1,4,3,6,2];
$diff = [];
for($i=0,$c=count($numbers);$i<$c;$i+=2)
{
    $diff[] = $numbers[$i]-$numbers[$i+1];
}
$sum = array_sum($diff);

Else, to just get the total and bypass the diff array:

$numbers = [2,1,4,3,6,2];
$total = 0;
for($i=0,$c=count($numbers);$i<$c;$i+=2)
{
    $total += $numbers[$i]-$numbers[$i+1];
}

Please check and see if this works for you.

    <?php        
    $sum=0;
    $number = array(2,1,4,3,6,2);
    for ($i=0;$i<=count($number);$i++) {
    if ($i%2 == 1 ) {  
        $sum = $sum + $number[$i-1] - $number[$i];
    }
    }

    print $sum;
    ?>

Try this one:

    $number = array(2,1,4,3,6,2);

    $diff = array();
    $v3 = 0;
    $i=1;
    foreach($number as $val){
    if ($i % 2 !== 0) {
        $v1 = $val;
    }
    if ($i % 2 === 0) {
        $v2 = $val;
        $diff[] = $v1-$v2;
        $v3+= $v1-$v2;       
        }
    $i++;
    }
    print $v3;//total value
    print_r($diff); //diff value array

I have got this far it gives the required solution.

$arr = array(2,1,4,3,6,2);
$temp = 0;
$diff = array();
foreach ($arr as $key => $value) {
    if($key % 2 == 0) {
        $temp = $value;
    }
    else {
        $diff[] = $temp - $value;
    }
}

print_R($diff);
print 'Total :' . array_sum($diff);

Note : Please update if any one knows any pre-defined function than can sorten this code.