0
下面是我的PHP代碼,目前,它顯示所有的錯誤等,但如果其中一個是正確的,它會提交表單,我如何更改我的代碼,以便如果1是不正確的是不與上次否則,如果提交PHP,不要提交,如果有錯誤
<?php
$cusMsg = "";
$fNameMsg = "";
if (isset($_POST["submit"])) {
$id = $_POST["custid"];
if(empty($id)) {
$cusMsg = '<span class="error"> Field was left empty</span>';
} else if(!is_numeric($id)) {
$cusMsg = '<span class="error"> Customer ID must be numeric</span>';
} else if(strlen($id) != 6) {
$cusMsg = '<span class="error"> Customer ID must be 6 digits long</span>';
} else {
return true;
}
}
if (isset($_POST["submit"])) {
$fName = $_POST["customerfname"];
$pattern = "/^[a-zA-Z-]+$/";
if(empty($fName)) {
$fNameMsg = '<span class="error"> Field was left empty</span>';
} else if(!preg_match($pattern, $fName)) {
$fNameMsg = '<span class="error"> First name must only containt letters and hyphens</span>';
} else if(strlen($fName) > 20) {
$fNameMsg = '<span class="error"> First name must not be longer than 20 characters</span>';
} else {
return true;
}
}
}
?>
手冊中去閱讀什麼'return'不 – 2016-04-24 05:31:45
@Dagon謝謝:)它仔細地閱讀起來,知道它的作用擺脫了對所有的人的陳述,然後把所有的代碼爲1 issest($ _ POST [「submit」]),它工作:) – Gajeel