Как найти нечетные числа в php

asked Oct 31, 2011 at 20:17

You were right in thinking mod was a good place to start. Here is an expression which will return true if $number is even, false if odd:

$number % 2 == 0

Works for every integer^PHP value, see as well Arithmetic Operators^PHP.

Example:

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

Output:

It’s even

Another option is a simple bit checking.

n & 1

for example:

if ( $num & 1 ) {
  //odd
} else {
  //even
}

answered Feb 5, 2012 at 23:10

Aaron W.Aaron W.

9,2442 gold badges33 silver badges45 bronze badges

While all of the answers are good and correct, simple solution in one line is:

$check = 9;

either:

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

or:

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

works very well.

answered May 9, 2016 at 15:58

(bool)($number & 1)

or

(bool)(~ $number & 1)

answered Jan 16, 2013 at 10:16

I did a bit of testing, and found that between mod, is_int and the &-operator, mod is the fastest, followed closely by the &-operator.
is_int is nearly 4 times slower than mod.

I used the following code for testing purposes:

$number = 13;

$before = microtime(true);
for ($i=0; $i<100000; $i++) {
    $test = ($number%2?true:false);
}
$after = microtime(true);

echo $after-$before." seconds mod<br>";

$before = microtime(true);
for ($i=0; $i<100000; $i++) {
    $test = (!is_int($number/2)?true:false);
}
$after = microtime(true);

echo $after-$before." seconds is_int<br>";

$before = microtime(true);
for ($i=0; $i<100000; $i++) {
    $test = ($number&1?true:false);
}
$after = microtime(true);

echo $after-$before." seconds & operator<br>";

The results I got were pretty consistent. Here’s a sample:

0.041879177093506 seconds mod
0.15969395637512 seconds is_int
0.044223070144653 seconds & operator

answered Aug 22, 2013 at 10:15

Another option is to check if the last digit is an even number :

$value = "1024";// A Number
$even = array(0, 2, 4, 6, 8);
if(in_array(substr($value, -1),$even)){
  // Even Number
}else{
  // Odd Number
}

Or to make it faster, use isset() instead of array_search :

$value = "1024";// A Number
$even = array(0 => 1, 2 => 1, 4 => 1, 6 => 1, 8 => 1);
if(isset($even[substr($value, -1)]){
  // Even Number
}else{
  // Odd Number
}

Or to make it more faster (beats mod operator at times) :

$even = array(0, 2, 4, 6, 8);
if(in_array(substr($number, -1),$even)){
  // Even Number
}else{
  // Odd Number
}

Here is the time test as a proof to my findings.

answered Oct 10, 2013 at 16:40

PHP is converting null and an empty string automatically to a zero. That happens with modulo as well. Therefor will the code

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

with value $number = » or $number = null
result in true.
I test it therefor somewhat more extended:

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;
}

This function returns 1 for any odd number, 0 for any even number and false when it is not a number. I always write === true or === false to let myself (and other programmers) know that the test is as intended.

answered Jun 1, 2015 at 8:12

All even numbers divided by 2 will result in an integer

$number = 4;
if(is_int($number/2))
{
   echo("Integer");
}
else
{
   echo("Not Integer");
}

answered Jul 5, 2012 at 19:39

Try this,

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

answered May 24, 2016 at 23:59

jxwdjxwd

1792 silver badges6 bronze badges

<?php
// Recursive function to check whether
// the number is Even or Odd 
function check($number){
    if($number == 0)
        return 1;
    else if($number == 1)
        return 0;
    else if($number<0)
        return check(-$number);
    else
        return check($number-2);        
}
  
// Check the number odd or even
$number = 35;
if(check($number))
    echo "Even";
else
    echo "Odd";
?>

So, the output will be Odd

answered Aug 20, 2021 at 10:52

Using Bit Manipulation:
In this method, you will find bit-wise AND of the number with 1. If the bit-wise AND is 1, then the number is odd, else even.

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>";

The output will be:

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
)

answered Nov 3, 2022 at 11:54

//checking even and odd
$num =14;

$even = ($num % 2 == 0);
$odd = ($num % 2 != 0);

if($even){
    echo "Number is even.";
} else {
    echo "Number is odd.";
}

answered Apr 19, 2018 at 6:55

Try this one with #Input field

<?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.";
    }
    }
?>

answered Apr 19, 2018 at 9:40

answered Feb 15, 2022 at 19:18