write form into database by using jquery

How can I send this form now to my .php page, where I will put the whole data into my database?

You have configured your form method as POST and you the attribute action = "index.php?content=create_tutorial"which is the endpoint or script that you are calling to do something with the data sent.

So, on submit, index.php should received the POST data using php global variable $_POST.

To simplify the example, the following form has one text input in a form using POST and action set as your endpoint.

HTML

<form method="post" action="index.php?content=create_tutorial">
  <input type="text" name="name"/><br>
  <input type="submit"/>
</form>

PHP

// index.php
<?php 
  $name = $_POST['name'];
  echo $name;
?>

PHP-MYSQL

// assuming connection to database is configured and setup done
$query = "INSERT INTO `user` (`name`) VALUES ($name)";
mysql_query($query);

Notes: - read more about security and validating user input here - take note of the attribute name value is the key to $_POST. - read about inserting database (assuming MySQL) here and deprecated mysql_query()

I have this code, but I dont know how to use it. What is success & data? Maybe someone of you can help me.

$.ajax() is a method to make HTTP requests asynchronously. Read more here on Wikipedia

• async = true, is to simply say "let me know when you are done, i am going to do other stuffs"
• async = false is to simply say "i'll wait for you before i do anything else"

By default, $.ajax() async flag is set to true. Because it implements the Promise interface,it provides you functions like

As of jQuery 1.5.1

• .done() when there is a successful response, this function will be called.
• .fail() when there is an error response, this function will be called.
• .. more in $.ajax() documentation

.php

// index.php
<?php
  return "ok";
?>

.js

var jqxhr = $.ajax( "example.php" )
     .done(function(results) {
       alert(results); // "ok"
     })
     .fail(function(error) {
       alert( error ); // "failed"
     })

Deprecation Notice: The jqXHR.success(), jqXHR.error(), and jqXHR.complete() callbacks are deprecated as of jQuery 1.8. To prepare your code for their eventual removal, use jqXHR.done(), jqXHR.fail(), and jqXHR.always() instead.

Hope this helps.

jQuery has already extensively documented the meaning and the various use cases of the $.ajax method here.

Basically, success is a function to be called if the request succeeds. It accepts a data argumentwhich is the data returned from the server (from your PHP backend). This is where you perform the actions you want to perform when the form data has been sent to the server and the server has responded. Note that the data from the server doesn't necessarily have to be used within the success function.