系统注释使用ajax jquery

Question

起初我想为弱小的英语道歉。我想使用jQuery ajax，但我不知道如何使用它。我想在没有预加载的情况下将数据表单发送到addcomment.php。

<?php
require_once('config/config.php');

if($_POST['sent']){
    $nick = $_POST['nick_ajax'];
    $content = $_POST['content_ajax'];
    $postID = $_POST['id_content_ajax'];

    $sql_comment = $PDO->prepare('insert into `comments` (`nick`,`content`,`postID`) values (:nick,:content,:id)');
    $sql_comment->execute(array(':nick' => $nick, ':content' => $content, ':id' => $postID));
    header('Location: home.php?success');
}
else {
    header('Location: home.php?fail');
    die();
}
?>

以下代码位于home.php

<form action="addcomment.php" method="POST">
    <input type="hidden" name="nick" id="nick_ajax" value="'.$_SESSION['user'].'" />
    <input type="text" style="max-width: 1100px;" id="content_ajax" class="form-control" placeholder="Write a comment" name="content" /><br />
    <input type="hidden" name="id" id="id_content_ajax" value="'.$article['id'].'" />
    <input type="submit" id="smb_ajax" class="btn btn-default" style="float:right; margin-top: -15px;"name="sent" value=SEND" />
</form>

现在我有了这段代码

$(document).ready(function(){
$("#smb_ajax").click(function(){
    var user = document.getElementById("nick_ajax").Value;
    var content = document.getElementById("content_ajax").Value;
    var id_cnt = document.getElementById("id_content_ajax").Value;
    var todo = 0;
    $.ajax({
    method: "POST",
    type: 'POST',
    url: "addcomment.php",
    data: { 
        'user': user,
        'content': content,
        'id_cnt':id_cnt
     },
     error: function() {
            alert('Error!');
        }
    })
});

}）;