我在PHP中编写表单验证代码,但我遇到了复选框验证问题。 浏览表单时,如果不选中该复选框,则会显示正确的错误消息。
但是,即使您选中了复选框,它仍然会显示相同的错误消息。
以下是目前的代码:
if (isset($_POST['firstname'], $_POST['lastname'], $_POST['address'], $_POST['email'], $_POST['password'])) {
$errors = array();
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$address = $_POST['address'];
$email = $_POST['email'];
$password = $_POST['password'];
if (empty($firstname)) {
$errors[] = 'First name can\'t be empty';
}
if (empty($lastname)) {
$errors[] = 'Last name can\'t be empty';
}
if (empty($address)) {
$errors[] = 'Address can\'t be empty';
}
if (filter_var($email, FILTER_VALIDATE_EMAIL) === FALSE) {
$errors[] = 'Please enter a valid email';
}
if (empty($password)) {
$errors[] = 'Password can\'t be empty';
}
}
if (!isset($checkbox)) {
$errors[] = 'Please agree to the privacy policy';
}
$sex = $_POST['sex'];
$age = $_POST['age'];
$checkbox = $_POST['checkbox'];
$_SESSION['validerrors'] = $errors;
$_SESSION['firstname'] = $firstname;
$_SESSION['lastname'] = $lastname;
$_SESSION['address'] = $address;
$_SESSION['email'] = $email;
$_SESSION['sex'] = $sex;
$_SESSION['age'] = $age;
$_SESSION['password'] = $password;
$_SESSION['checkbox'] = $checkbox;
if (!empty($errors)) {
header('Location: index.php');
} else {
header('Location: writevalues.php');
}
上述代码中的其他所有内容都运行正常,但我无法找到任何有用的答案来验证复选框验证情况。提前致谢!
答案 0 :(得分:5)
您正在调用此代码:
if (!isset($checkbox)) {
$errors[] = 'Please agree to the privacy policy';
}
在此之前:
$checkbox = $_POST['checkbox'];
所以当然isset($checkbox)
会返回false
,因为它没有在那时设置!
您可以将if语句更改为:
if(!isset($_POST['checkbox'])){ ...
或者将行$checkbox = $_POST['checkbox'];
移到if语句之上。