Updating Database from Javascript by calling PHP script

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>

Do u want to pass data from the js to php ? and php will pass the data to the db ? — underscore
– underscore, Commented May 1, 2013 at 6:10
An AJAX call is probably best... If you don't want it to return anything, then you can just return some garbage and ignore it. — Radu Murzea
– Radu Murzea, Commented May 1, 2013 at 6:14
you need something like ajax stackoverflow.com/questions/12026600/… — M Khalid Junaid
– M Khalid Junaid, Commented May 1, 2013 at 6:34

Yadav Chetan · Accepted Answer · 2013-05-01 07:00:59Z

3

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>

answered May 1, 2013 at 7:00

Yadav Chetan

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4 Comments

Jared Over a year ago

You might want to do some validation to make sure it was actually successful instead of assuming that a successful AJAX call was a successful like. That's up to the OP though.

Yadav Chetan Over a year ago

@Jared yes that can be checked through the response of LikeUnlike.php i just give the way he can do... i should not give all the code right?

Jared Over a year ago

Yeah the server part of it is for another question if OP can't do it themselves.

Tom Hanson Over a year ago

So am I right in saying to post to the php file I would use data:{}

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