Ajax 验证在 html 元素中复制 html 页面

Question

我使用 Ajax 的 PHP 用户名验证在 htmldiv 元素中复制了我的 html 页面(这是为了显示 ajax 错误(元素。我尝试了一些解决方案，谷歌它找不到任何其他解决方案。也许问题在于 $_POST，但我也在 php 中将它们分开(所有输入验证(。

这是PHP代码

<?php 
if(isset($_POST['username'])){
//username validation
$username = $_POST['username'];
if (! $user->isValidUsername($username)){
$infoun[] = 'Your username has at least 6 alphanumeric characters';
} else {
$stmt = $db->prepare('SELECT username FROM members WHERE username = :username');
$stmt->execute(array(':username' => $username));
$row = $stmt->fetch(PDO::FETCH_ASSOC);
if (! empty($row['username'])){
$errorun[] = 'This username is already in use';
}
}
}

if(isset($_POST['fullname'])){
//fullname validation
$fullname = $_POST['fullname'];
if (! $user->isValidFullname($fullname)){
$infofn[] = 'Your name must be alphabetical characters';
}   
}
if(isset($_POST['password'])){
if (strlen($_POST['password']) < 6){
$warningpw[] = 'Your password must be at least 6 characters long';
}   
}
if(isset($_POST['email'])){
//email validation
$email = htmlspecialchars_decode($_POST['email'], ENT_QUOTES);
if (! filter_var($email, FILTER_VALIDATE_EMAIL)){
$warningm[] = 'Please enter a valid email address';
} else {
$stmt = $db->prepare('SELECT email FROM members WHERE email = :email');
$stmt->execute(array(':email' => $email));
$row = $stmt->fetch(PDO::FETCH_ASSOC);
if (! empty($row['email'])){
$errorm[] = 'This email is already in use';
}
}
}
?>

这是Javascript

<script type="text/javascript">
$(document).ready(function(){

$("#username").keyup(function(event){
event.preventDefault();
var username = $(this).val().trim();
if(username.length >= 3){
$.ajax({
url: 'register.php',
type: 'post',
data: {username:username},
success: function(response){
// Show response
$("#uname_response").html(response);
}
});
}else{
$("#uname_response").html("");
}
});     
});
</script>
<input type="text" name="username" id="username" class="form-control form-control-user" placeholder="Kullanıcı Adınız" value="<?php if(isset($error)){ echo htmlspecialchars($_POST['username'], ENT_QUOTES); } ?>" tabindex="2" required>
<div id="uname_response" ></div>

这是屏幕截图：

表单重复屏幕截图

Answer 1

PHP 文件中唯一的代码应该在<?php ?>标记内。您需要将PHP代码分离到另一个文件中。