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
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.";
}
}
?>
其他回答
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
所有偶数除以2都得整数
$number = 4;
if(is_int($number/2))
{
echo("Integer");
}
else
{
echo("Not Integer");
}
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。
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.";
}
}
?>
//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.";
}