i want to count number of even,number of odd in php - php

Question

I want to count number of even,number of odd.If there 2 even ,i count only one even,If there 2 odd i count
only one number of odd
<?php
$myarray = array(5,5,0,1,2,1,1,6,1);
for ($i = 0; $i < count($myarray); $i++) {
echo "Index ", $i, ", value ", $myarray[$i], ": ";
if ($myarray[$i] % 2 == 0) {
echo "even\n";
}
else {
echo "odd\n";
}
}
?>
input =[5,5,0,1,2,1,1,6,1]
output = 5 :1(total(5) odd one),
1:2 (total(1) odd two)

Answer 1

Use two variables that keep track of your odds and evens when you loop through your array.
$my_array = array( 5, 5, 0, 1, 2, 1, 1, 6, 1 );
$odd = $even = 0;
foreach ( $my_array as $number ) {
$number % 2 == 0 ? $even++ : $odd++;
}
printf( 'Odd: %s | Even: %s', $odd, $even );

Answer 2

<?php
$num = array(23,12,11,9,6,7,4,5,3);
$n=count($num);
$even = 0;
$odd = 0;
for( $i = 0 ; $i < $n; $i++)
{
if ($num[$i] & 1 == 1)
$odd ++ ;
else
$even ++ ;
}
echo "Number of even elements = $even
and Number of odd elements = $odd" ;
?>

Answer 3

Create counters $evens and odds. Then Loop through elements in array check if number is odd
using Ternary Operator (condition) ? true : false;, and $num & 1 return true if number is odd.
$nums = array(5,5,0,1,2,1,1,6,1);
$odds = $evens = 0;
foreach ($nums as $num) {
$num & 1 ? ++$odds : ++$evens;
}
echo "Odds: $odds, Evens: $evens"; //Odds: 6, Evens: 3

Answer 4

Just count only number of odds, Because If we count number of Odds in array, It would be easy to know number of Evens by subtract number of elements in array with number of Odds
$nums = array(5,5,0,1,2,1,1,6,1);
$odds = 0;
foreach ($nums as $num) {
$odds += $num % 2;
}
$evens = count($nums)-$odds;
echo "Odds: $odds, Evens: $evens"; //Odds: 6, Evens: 3