如何插入ajax方法,通过按下按钮将用户ID发布到mysql,刷新页面后,按钮上显示相同的文本

-------------------------HTML code----------------------------
<body class="bdy">

    <?php include "includes/head.php"; ?>




<?php

//include "includes/connection.php";
 $us=$_GET['Uid']; (user id to post)
 echo '<div class="container">
    <button class="btn followButton follow" rel="6">Follow</button>
</div>';


include "includes/uprofile1.php";
include "includes/footer.php";

?>


<script src="style/javascript/follow.js"></script>

</body>

-------------------jquery code-----------------------
$(document).ready(function(){

$('button.followButton').on('click', function(e){
    e.preventDefault();
    $button = $(this);


    if($button.hasClass('following')){

        //$.ajax(); Do Unfollow

//this is only the button code i need help in adding the ajax call....

        $button.removeClass('following');
        $button.removeClass('unfollow');
        $button.text('Follow');
    } else {
     var data=$(this).attr('Uid');
     $.ajax({
        url:"follow.php",
        type:"post",
        data:data,


     });
    $button.addClass('following');
        $button.text('Following');

    }
});

$('button.followButton').hover(function(){
     $button = $(this);
    if($button.hasClass('following')){
        $button.addClass('unfollow');
        $button.text('Unfollow');
    }
}, function(){
    if($button.hasClass('following')){
        $button.removeClass('unfollow');
        $button.text('Following');
    }
});

    });

please someone help me in inserting the ajax post, to post the user id to mysql .i don't understand how to make the ajax call and all.

Since you provided no errors i will guess..

 $.ajax({
    url:"follow.php",
    type:"post",
    data:data
 });

Had to remove the extra comma after data:data