这个AJAX调用代码缺少什么?

Question

我很抱歉没有提供完整的上下文代码，我是非常新的。下面是sign .php文件的代码:

<?php
session_start();
include('connection.php');

$missingUsername='<p><strong>Please enter a username</strong></p>';
$missingEmail='<p><strong>Please enter your email address</strong></p>';
$InvalidEmail='<p><strong>Please enter a valid email address</strong></p>';
$missingPassword='<p><strong>Please enter a password</strong></p>';
$InvalidPassword='<p><strong>Your password should be at least 6 characters long and include one capital letter and one number</strong></p>';
$differentPassword='<p><strong>Passwords don't match</strong></p>';
$missingPassword2='<p><strong>Please confirm your password</strong></p>';

if(empty($_POST["username"])){
$errors .= $missingUsername;
}else{
$username = filter_var($_POST["username"], FILTER_SANITIZE_STRING);
}

if(empty($_POST["email"])){
$errors .= $missingEmail;
}else{
$email = filter_var($_POST["email"], FILTER_SANITIZE_EMAIL);
if(!filter_var($email, FILTER_VALIDATE_EMAIL)){
$errors .= $InvalidEmail; 
}
}

if(empty($_POST["password"])){
$errors .= $missingPassword;
}elseif(!(strlen($_POST["password"])>6 and preg_match('/[A-Z]/',$_POST["password"]) and preg_match('/[0-9]/',$_POST["password"]))){
$errors .= $InvalidPassword;
}else{
$password = filter_var($_POST["password"], FILTER_SANITIZE_STRING);
if(empty($_POST["password2"])){
$errors .= $missingPassword2;
}else{
$password2 = filter_var($_POST["password2"], FILTER_SANITIZE_STRING);
if($password !== $password2){
$errors .= $differentPassword;
}
}
}

if($errors){
$resultMessage = '<div class="alert alert-danger">' . $errors .'</div>'
echo $resultMessage;
}

$username = mysqli_real_escape_string($link, $username);
$email = mysqli_real_escape_string($link, $email);
$password = mysqli_real_escape_string($link, $password);

$password = hash('sha256', $password);
$sql = "SELECT * FROM users WHERE username = '$username'";
$result = mysqli_query($link, $sql);
if(!$result){
echo '<div class="alert alert-danger">Error running the query!</div>';
exit;
}
$results = mysqli_num_rows($result);
if($results){
echo '<div class="alert alert-danger">That username is already registered. Do you want to log in?</div>';
exit;
}
$sql = "SELECT * FROM users WHERE email = '$email'";
$result = mysqli_query($link, $sql);
if(!$result){
echo '<div class="alert alert-danger">Error running the query!</div>';
exit;
}
$results = mysqli_num_rows($result);
if($results){
echo '<div class="alert alert-danger">That email is already registered. Do you want to log in?</div>';
exit;
}

$activationKey = bin2hex(openssl_random_pseudo_bytes(16));

$sql = "INSERT INTO users ('username', 'email', 'password', 'activation') VALUES ('$username', '$email', '$password', '$activationKey')";
$result = mysqli_query($link, $sql);
if(!$result){
echo '<div class="alert alert-danger">There was an error inserting the user details in the database!</div>';
exit;
}

$message = "Please click on this link to activate your account:nn";
$message .= "http://studenttest.host20.uk/activate.php?email=" . urlencode($email) . "&key=$activationKey";
if(mail($email, 'Confirm your Registration', $message, 'From:'.'[e-mail]')){
echo "<div class='alert alert-success'>Thank you for registration! Confirmation email has been sent to $email. Please click on the activation link to activate your account.</div>";
}
?>

这里再次是JS代码块，我正在制作一个适当的注册表单的广泛目标内工作:

$("#signupform").submit(function(event){
event.preventDefault();
var datatopost = $(this).serializeArray();
console.log(datatopost);
$.ajax({
url: "signup.php",
type: "POST",
data: datatopost,
success: function(data){
if(data){
$("#signupmessage").html(data);
}
},
error: function(){
$("#signupmessage").html("<div class='alert alert-danger'>There was an error with the Ajax call. Please try again later.</div>");
}
});
});

由于某种原因，我一直得到AJAX错误，而不是为用户名输入、密码输入等设置的典型错误消息。它被链接到正确的文件(放在我的主index.php文件的script标签中)和其他所有文件。

发生了什么事?

Answer 1

你可能有一个错误在你的注册。php页面，所以它会更好，如果你把整个代码，以便找到问题并解决它。

我做了一个小例子，几乎与你的代码相同，它工作得很好:

<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/bootstrap@4.5.3/dist/css/bootstrap.min.css" >

<div id="signupmessage"></div>
<form id="signupform" action="" method="POST">
<input type="text" name="firstname" placeholder="Enter your first name" /><br /><br />
<input type="text" name="lastname" placeholder="Enter your last name" /><br /><br />
<input type="submit" value="submit">
</form>

<script src="https://code.jquery.com/jquery-3.6.0.min.js"></script>
<script type='text/javascript'>
$("#signupform").submit(function(event){
event.preventDefault();
var datatopost = $(this).serializeArray();
console.log(datatopost);
$.ajax({
url: "signup.php",
type: "POST",
data: datatopost,
success: function(data){
$("#signupmessage").html(data);
},
error: function(data){
$("#signupmessage").html(data);
}
});
});
</script>

--------- sign .php -----------

<?php
if(isset($_POST['firstname']) && isset($_POST['lastname'])){

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];

echo $firstname."<br />";
echo $lastname;
}
else{
echo "<div class='alert alert-danger'>There was an error with the Ajax call. Please try again later.</div>";
}

?>