$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,

$number = 10;
 switch ($number%2)
 {
 case 0:
 echo "It's even";
 break;
 default:
 echo "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。

PHP自动将null和空字符串转换为0。对模也是这样。因此，代码将

$number % 2 == 0 or !($number & 1)

$number = "或$number = null 结果为true。 因此，我对它进行了更广泛的测试:

function testEven($pArg){
    if(is_int($pArg) === true){
        $p = ($pArg % 2);
        if($p === 0){
            print "The input '".$pArg."' is even.<br>";
        }else{
            print "The input '".$pArg."' is odd.<br>";
        }
    }else{
        print "The input '".$pArg."' is not a number.<br>";
    }
}

The print is there for testing purposes, hence in practice it becomes:
function testEven($pArg){
    if(is_int($pArg)=== true){
        return $pArg%2;
    }
    return false;
}

这个函数对于任何奇数返回1，对于任何偶数返回0，当它不是数字时返回false。我总是写=== true或=== false来让我自己(和其他程序员)知道测试是预期的。

使用位操作: 在此方法中，您将找到带有1的数字的逐位与。如果位与为1，则该数字为奇数，否则为偶数。

function odd($var){
    return $var & 1;
}
function even($var){
    return !($var&1);
}
$arr = range(1,10);

echo "Odd Numbers: ";
echo "<pre>";
    print_r(array_filter($arr2, 'odd'));
echo "<pre>";

echo "<br>Even Numbers: ";
echo "<pre>";
    print_r(array_filter($arr2, 'even'));
echo "</pre>";

输出将是:

Odd Numbers:
Array
(
    [0] => 1
    [2] => 3
    [4] => 5
    [6] => 7
    [8] => 9
)

Even Numbers: 
Array
(
    [1] => 2
    [3] => 4
    [5] => 6
    [7] => 8
    [9] => 10
)

$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";
}