在PHP中找出一个数字/变量是奇数还是偶数的最简单最基本的方法是什么? 这和mod有关吗?
我试过一些剧本,但是…谷歌目前没有发送。
在PHP中找出一个数字/变量是奇数还是偶数的最简单最基本的方法是什么? 这和mod有关吗?
我试过一些剧本,但是…谷歌目前没有发送。
当前回答
//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.";
}
其他回答
$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";
}
所有偶数除以2都得整数
$number = 4;
if(is_int($number/2))
{
echo("Integer");
}
else
{
echo("Not Integer");
}
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.";
}
}
?>
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($arr2, 'even'));
echo "</pre>";
输出将是:
Odd Numbers:
Array
(
[0] => 1
[2] => 3
[4] => 5
[6] => 7
[8] => 9
)
Even Numbers:
Array
(
[1] => 2
[3] => 4
[5] => 6
[7] => 8
[9] => 10
)
<?php
// Recursive function to check whether
// the number is Even or Odd
function check($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";
?>
因此,输出将是Odd