how to insert the ajax method to post the user id to mysql by pressing the button and also after refreshing the page the same text appears on button

Question

-------------------------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.

Answer 1

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

how to insert the ajax method to post the user id to mysql by pressing the button and also after refreshing the page the same text appears on button

Question

1 answers

solution1
0 2014-08-23 13:03:26

how to insert the ajax method to post the user id to mysql by pressing the button and also after refreshing the page the same text appears on button

Question

1 answers

solution1 0 2014-08-23 13:03:26

solution1
0 2014-08-23 13:03:26