我有X = 10,我想检查它是否甚至然后'赢'与0-9的值比较,如果x = 1,3,5,7,9 =赢,否则'输',但不想使用% 2 == 0
if(X=1,3,5,7,9){
echo "win";
}
谢谢
如何使用in_array
if (in_array($x, [1,3,5,7,9]) {
echo "win";
}
这里是正确的方式$ x%2为偶数= 0,对于奇数= 1,当奇数值到来时它将回显win,在数组中你必须在数组中定义奇数值。
if ($x%2) {
echo "win";
}
对https://www.geeksforgeeks.org/php-check-number-even-odd/的信贷
试试这个Bitwise AND
方法and its is NOT USING %2 == 0
function checkIsEvenOrOdd($number)
{
// One
$one = 1;
// Bitwise AND
$bitwiseAnd = $number & $one;
if($bitwiseAnd != 1)
{
echo "Even";
}
else{
echo "Odd";
}
}
echo checkIsEvenOrOdd('2');
echo checkIsEvenOrOdd('1');
echo checkIsEvenOrOdd('97');
希望它有所帮助