I have this code, which is a form to leave comments on my website. I would like to know if there is any way to get that by adding something else code I can get a notice to my email when someone leaves a comment on the web. The code that I leave works well, it does not have failures I only put it so that you can see what I am using. Thanks in advance.
<div class="container">
<form method="POST" id="comment_form">
<div class="form-group">
<input type="text" name="comment_name" id="comment_name" class="form-
control" placeholder="Enter Name" />
</div>
<div class="form-group">
<textarea name="comment_content" id="comment_content" class="form-control"
placeholder="Enter Comment" rows="5"></textarea>
</div>
<div class="form-group">
<input type="hidden" name="comment_id" id="comment_id" value="0" />
<input type="submit" name="submit" id="submit" class="btn btn-info"
value="Submit" />
</div>
add_comment.php:
<?php
//add_comment.php
$connect = new PDO('mysql:host=);
$error = '';
$comment_name = '';
$comment_content = '';
if(empty($_POST["comment_name"]))
{
$error .= '<p class="text-danger">Name is required</p>';
}
else
{
$comment_name = $_POST["comment_name"];
}
if(empty($_POST["comment_content"]))
{
$error .= '<p class="text-danger">Comment is required</p>';
}
else
{
$comment_content = $_POST["comment_content"];
}
if($error == '')
{
$query = "
INSERT INTO tbl_comment
(parent_comment_id, comment, comment_sender_name)
VALUES (:parent_comment_id, :comment, :comment_sender_name)
";
$statement = $connect->prepare($query);
$statement->execute(
array(
':parent_comment_id' => $_POST["comment_id"],
':comment' => $comment_content,
':comment_sender_name' => $comment_name
)
);
$error = '<label class="text-success">Comment Added</label>';
}
$data = array(
'error' => $error
);
echo json_encode($data);
?>