how to insert JavaScript variably in database

Data has to be an object. The property of the object is what you fetch in your $_POST in your PHP script.

      $('#add').click(function(){
       var num = 0;

        $.ajax({
            url: 'php/insertlevel.php',
            type: 'post',
            data: {num: num},
        });
   });

Your PHP file looks good (assuming $con is the connection). There is only one thing you missed. The single quotes around the value.

<?php 
    include "connect.php";
    $val=$_POST['num'];
    echo "i am called dont worry";
    $sql="INSERT INTO `level` (`level`) VALUES ('$val')"; //NOTICE, YOU SEE THE SINGLE QUOTES AROUND THE VAL VARIABLE?
    $insert=mysqli_query($con,$sql);

    if ($insert) {
        echo "Data insert";
    } else{
        echo "not insert";
      } 
 ?>

This should do the trick for you.

However, after you have "mastered" this, I highly recommend you to look at PDO or MYSQLI and use the prepared statements. This will ensure you are safe against SQL injections.

I recommend you to look at the PHP manual, but (atleast, for me it was) w3schools explains it in a more of a newbie way. It is however a highly discouraged website when you grow as a developer and actually have to implement for a live environment.

https://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php https://www.w3schools.com/php/php_mysql_prepared_statements.asp

The data option can contain either a query string of the form key1=value1&key2=value2, or an object of the form {key1: 'value1', key2: 'value2'}

Try to change your code to this:

var data = {num: 0};

  $('#add').click(function(){

    $.ajax({

        url: 'php/insertlevel.php',
        method: 'POST',
        data: data,


    });

    });

or if you like to send it as a string, change the data variable to data = "num=0" instead of num = 0; The way you doing, with num=0, the data property is set to 0 {data: 0}, you need to send the data object like this: {data: {num: 0}}

Reference: https://api.jquery.com/jquery.ajax/

Please note you should never trust the user input, it's insecure because doing that you are vulnerable to SQL injections attacks. Your SQL query must be parametrized like that:

https://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php