Ajax post response returns empy if database connection fails

I am building a PHP project that has a simple form with 3 fields. This form is connected with a MySQL table named "article". When I submit the form, with the help of jQuery and Ajax I display a notification message: eg "Submitted successfully" on success submission or "Something went wrong" on fail submission. My problem is that this notification message works fine on successful submission but if there is an error - For example if database connection fails - then the notification message is empty and it doesn't display the error message! I added the console.log(error); in my code so I can check the console for errors but there are no errors displayed there. What I 'm doing wrong?

Here is the PHP code I wrote for submit the form in the DB:

<?php
require __DIR__ . '/../config_db.php';
if ($_SERVER["REQUEST_METHOD"] == "POST") {
// Retrieve the submitted values
$article = $_POST["article"];
$author = $_POST["author"];
$tags = $_POST["tags"];
// Create a new database connection
$connection = new mysqli($servername, $username, $password, $dbname);
// Check the connection
if ($connection->connect_error) {
$response = array(
"status" => "error",
"message" => "Error: Failed to connect to the database."
);
} else {
// Prepare the statement
$stmt = $connection->prepare("INSERT INTO article (article, author, tags) VALUES (?, ?, ?)");
// Bind parameters to the statement
$stmt->bind_param("sss", $article, $author, $tags);
if ($stmt->execute()) {
$response = array(
"status" => "success",
"message" => "Article submitted successfully!"
);
} else {
$response = array(
"status" => "error",
"message" => "Error: " . $stmt->error
);
}
// Close the statement
$stmt->close();
// Close the database connection
$connection->close();
}
// Send the JSON response
header('Content-Type: application/json');
echo json_encode($response);
}
?>

And here's the html form with ajax:

 <form method="post" action="submit_article.php" id="articleForm">
<div class="field">
<div class="control">
<textarea class="textarea" id="article" name="article" placeholder="Enter the article..." required></textarea>
</div>
</div>
<div class="field">
<div class="control">
<input class="input" type="text" id="author" name="author" placeholder="Author name..." required>
</div>
</div>
<div class="field">
<div class="control">
<input class="input" type="text" id="tags" name="tags" placeholder="Add some tags...">
</div>
</div>
<div class="field is-grouped">
<div class="control">
<input class="button is-link" type="submit" value="Submit">
</div>
<div class="control">
<input class="button is-link is-light" type="button" value="Cancel" onclick="window.location.href='index.php'">
</div>
</div>
</form>
<div id="notification" class="notification is-primary"></div>
<script>
$(document).ready(function() {
$("#articleForm").submit(function(e) {
e.preventDefault(); // Prevent form submission
// Send the form data via AJAX
$.ajax({
url: "submit_article.php",
type: "POST",
data: $(this).serialize(),
success: function(response) {
// Display the notification message
var notification = $("#notification");
notification.text(response['message']);
notification.fadeIn().delay(3000).fadeOut();
// Clear the form
$("#article").val("");
$("#author").val("");
$("#tags").val("");
},
error: function(xhr, status, error) {
// Display the error message
var notification = $("#notification");
notification.text(response['message']);
notification.fadeIn().delay(3000).fadeOut();
console.log(error); // Log any errors to the console
}
});
return false; // Prevent default form submission
});
});
</script>

using $connection->connect_error will give you an error for well known errors, such as wrong user / password / permission, etc.

Having a real connection issue will cause new mysqli($servername, $username, $password, $dbname); to throw an exception. Hence your script is simply interrupted by the generic exception handler and the error-response you are generating due to $connection->connect_error will never be generated.

To cover that, you need to use a try/catch-block and generate a proper response there.

Update:
With mysqli_report(MYSQLI_REPORT_STRICT); you can force mysqli to throw exceptions for well known errors as well - then you can ommit the additional If check and handle everything from within the catch-block.