你认为mod是个不错的开始，这是对的。下面是一个表达式，如果$number为偶数则返回true，如果为奇数则返回false:

$number % 2 == 0

适用于所有integerPHP值，参见算术操作符sphp。

例子:

$number = 20;
if ($number % 2 == 0) {
  print "It's even";
}

输出:

它甚至

两个简单的位函数，返回0表示False，返回1表示True。

# is_odd: 1 for odd , 0 for even
odd = number & 1

# is_even: 1 for even , 0 for odd
even = number & 1 ^ 1

$before = microtime(true);

$n = 1000;  
$numbers = range(1,$n);

$cube_numbers = array_map('cube',$numbers);

function cube($n){      
    $msg ='even';       
    if($n%2 !=0){
        $msg = 'odd';
    }               
    return "The Number is $n is ".$msg;
}

foreach($cube_numbers as $cube){
    echo $cube . "<br/>";
}

$after = microtime(true);

echo $after-$before. 'seconds';

试试这个带有#Input字段的

<?php
    //checking even and odd
    echo '<form action="" method="post">';
    echo "<input type='text' name='num'>\n";
    echo "<button type='submit' name='submit'>Check</button>\n";
    echo "</form>";

    $num = 0;
    if ($_SERVER["REQUEST_METHOD"] == "POST") {
      if (empty($_POST["num"])) {
        $numErr = "<span style ='color: red;'>Number is required.</span>";
        echo $numErr;
        die();
      } else {
          $num = $_POST["num"];
      }


    $even = ($num % 2 == 0);
    $odd = ($num % 2 != 0);
    if ($num > 0){
        if($even){
            echo "Number is even.";
        } else {
            echo "Number is odd.";
        }
    } else {
        echo "Not a number.";
    }
    }
?>

$number %2 = 1如果是奇数…所以不用用not even…

$number = 27;

if ($number % 2 == 1) {
  print "It's odd";
}

虽然所有的答案都是正确的，但简单的解决方法是:

$check = 9;

:

echo ($check & 1 ? 'Odd' : 'Even');

or:

echo ($check % 2 ? 'Odd' : 'Even');

工作得很好。