用PHP脚本交流javascript

Question

I am trying to send data from Javascript to a php script. I am new to PHP,javascript and JSON etc. I am tryin to use jQuery Upvote plugin. I am reading the documentation from readme.md here. However, I find it too vague for my understanding.

I have added the following code in HTML body. The plugin is working fine in front-end.

<div id="topic" class="upvote">
    <a class="upvote"></a>
    <span class="count">0</span>
    <a class="downvote"></a>
    <a class="star"></a>
</div>
<div id="output"> </div>
<script language="javascript">
var callback = function(data) {
    $.ajax({
        url: '/vote',
        type: 'post',
        data: { id: data.id, up: data.upvoted, down: data.downvoted, star: data.starred }
    });
};
$('#topic').upvote({id: 123, callback: callback});
</script>

Can someone please tell how should I write the php counterpart so as to get the information about change of state of plugin? I think it is related to json maybe but I am stuck here..

EDIT: As per the suggestion, I have modified a bit of code and wrote PHP script. In above code I have changed URL to my PHP script url: 'voter.php'.

This is the PHP code (in brief) :

<?php require_once('Connections/conn.php'); ?>
mysql_select_db($database_conn, $conn);
if ($_POST['up']!=0){
$query_categories = "UPDATE user SET score=1 WHERE u_id =5";
$categories = mysql_query($query_categories, $conn) or die(mysql_error());
}
else if ($_POST['down']!=0) {
    $query_categories = "UPDATE user SET score=-1 WHERE u_id =5";
$categories = mysql_query($query_categories, $conn) or die(mysql_error());

}

The database connection is working fine, but still the database values are not updated by above code..

Answer 1

Your script sends AJAX Post request to the php script. You can access those post variables like this, using $_POST[] array:

$id = $_POST['id'];
$up = $_POST['up'];
$down = $_POST['down'];
$star = $_POST['star'];

Then you can populate the mysql table, etc.