Updating Database from Javascript by calling PHP script - php

Question

Ive been trying to get this for age
I am building a website which has an activity wall. I have the whole thing working except the like and unlike buttons. I have them currently just showing a text box I like or I don't like
<a href='#' onclick='like()'>Like</a>
or
<a href='#' onclick='unlike()'>Unlike</a>
Now these call these scripts
<script>
function like()
{
alert("I like");
document.getElementById("p1").innerHTML="<a href='#' onclick='unlike();'>Unlike</a> | 1 Life<hr/>";
}
function unlike()
{
alert("Dont like anymore");
document.getElementById("p1").innerHTML="<a href='#' onclick='like();'>Like</a> | 0 Lifes<hr/>";
}
</script>
I have a php script that connects to the database and adds or removes the like which I know works. I don't need to return anything to the javascript but how do I call that php script giving the postID and username??
Ok so using this I have modified it a little. But it still doesnt work :S
==LikeUnlink.php==
<?php
include 'phpScripts/OpenConnection.php';
$mode = $_GET['mode'];
$username = $_GET['username'];
$postID = $_GET['LikeID'];
echo $username."<br/>";
echo $mode."<br/>";
echo $postID."<br/>";
if($mode == 0)
{
$query = "INSERT INTO tbl1Ups (Username, ActivityID) VALUES ('$username', $postID)";
}
else
{
$query = "DELETE FROM `tbl1Ups` WHERE Username='$username' AND ActivityID=$postID";
}
$result = mysql_query($query) or die(mysql_error());
==MembersHome.php==
<script type="text/javascript">
function process(LikeId, Username, currentLikes, likeUnlike)
{
//your validation code
$.ajax(
{
type: 'GET',
url: LikeUnlike.php, //file where like unlike status change in database
data:{like:LikeId, username:Username, mode:likeUnlike},
success: function(data)
{
alert('It Works');
}
} );
}
</script>
==Like link==
<a href='Javascript:void(0)' onclick='process(".$row['ID'].", \"".$username."\", ".$likes.", 0);'>$linkText</a>

Answer 1

try this
<script type="text/javascript">
function process(LikeId) {
//your validation code
$.ajax( {
type: 'POST',
url: LikeUnlike.php, //file where like unlike status change in database
data:{like:LikeId},
success: function(data) {
//code you want to do after successfull process
}
} );
}
</script>
make changes in html
<a href='Javascript:void(0)' onclick='process(1)'>Like</a>
<a href='Javascript:void(0)' onclick='process(0)'>Unlike</a>