我正在使用此代码来检查数字是偶数还是奇数。我正在学习 PHP，当我运行这段代码时，它给出了一个异常错误。
<html>
<body>
<head>
<title>Judging even and odd numbers</title>
</head>
<form method = "post" action="EAO.php" >
<font size = "20">Please enter a number to check if it is even or odd:</font>
<input type = "text" name = "number" />
<input type = "hidden" name="checker" value="true" />
<input type = "submit" value = "submit" />
</form>
<?PHP
if (isset($_POST['checker'])) {
$number = $_POST['number'];
if ($number % 2 == 0 ) {
echo "The number is Even";
}
if($number % 2 == 1 ) {
echo "The number is odd";
}
if ($number == "") {
echo "Please enter a number";
}
}
?>
</body>
</html>
当你输入一个数字时效果很好，但是当你提交空的表单时，它说“请输入一个数字。数字是偶数”。如果我做错了什么，请帮助我更正。如果我的代码不符合判断数字的标准，请编写更好的代码。谢谢。
最佳答案
只要颠倒你的逻辑就会为你创造奇迹。您必须始终使用 if () {....} elseif () {....}，以便在给定时间只运行一个逻辑。
<html>
<head>
<title>Judging even and odd numbers</title>
</head>
<body>
<form method = "post" action="EAO.php" >
<font size = "20">Please enter a number to check if it is even or odd:</font>
<input type = "text" name = "number" />
<input type = "hidden" name="checker" value="true" />
<input type = "submit" value = "submit" />
</form>
<?PHP
if (isset($_POST['checker'])) {
$number = $_POST['number'];
if ($number == "") {
echo "Please enter a number";
}
else if ($number % 2 == 0 ) {
echo "The number is Even";
}
else if($number % 2 == 1 ) {
echo "The number is odd";
}
}
?>
</body>
</html>
关于PHP偶数或奇数脚本，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32665853/